Q1(i): Find the reference angle of $110^\circ$.

Step 1: Identify the quadrant of the angle:

The given angle $\theta = 110^\circ$ lies in the second quadrant ($90^\circ < 110^\circ < 180^\circ$).

Step 2: Apply the reference angle formula for Quadrant II:

$$\theta_{\text{ref}} = 180^\circ - \theta$$

$$\theta_{\text{ref}} = 180^\circ - 110^\circ = 70^\circ$$

Final Answer:

$$\mathbf{\theta_{\text{ref}} = 70^\circ}$$

Q1(ii): Find the reference angle of $138^\circ$.

Step 1: Identify the quadrant of the angle:

The angle $\theta = 138^\circ$ lies in the second quadrant ($90^\circ < 138^\circ < 180^\circ$).

Step 2: Apply the reference angle formula for Quadrant II:

$$\theta_{\text{ref}} = 180^\circ - \theta$$

$$\theta_{\text{ref}} = 180^\circ - 138^\circ = 42^\circ$$

Final Answer:

$$\mathbf{\theta_{\text{ref}} = 42^\circ}$$

Q1(iii): Find the reference angle of $125^\circ$.

Step 1: Identify the quadrant of the angle:

The angle $\theta = 125^\circ$ lies in the second quadrant ($90^\circ < 125^\circ < 180^\circ$).

Step 2: Apply the reference angle formula for Quadrant II:

$$\theta_{\text{ref}} = 180^\circ - \theta$$

$$\theta_{\text{ref}} = 180^\circ - 125^\circ = 55^\circ$$

Final Answer:

$$\mathbf{\theta_{\text{ref}} = 55^\circ}$$

Q1(iv): Find the reference angle of $142^\circ$.

Step 1: Identify the quadrant of the angle:

The angle $\theta = 142^\circ$ lies in the second quadrant ($90^\circ < 142^\circ < 180^\circ$).

Step 2: Apply the reference angle formula for Quadrant II:

$$\theta_{\text{ref}} = 180^\circ - \theta$$

$$\theta_{\text{ref}} = 180^\circ - 142^\circ = 38^\circ$$

Final Answer:

$$\mathbf{\theta_{\text{ref}} = 38^\circ}$$

Q2: If $\theta = 36^\circ$, find the equivalent angle $\alpha$ in the second quadrant for which $\sin \alpha = \sin \theta$.

Step 1: Understand the trigonometric relation for sine in Quadrant II:

In the unit circle, the sine function is positive in both the first and second quadrants. The relationship between an angle $\theta$ and its supplementary angle $\alpha$ in the second quadrant is:

$$\sin(180^\circ - \theta) = \sin \theta$$

Step 2: Calculate the angle $\alpha$:

$$\alpha = 180^\circ - \theta = 180^\circ - 36^\circ = 144^\circ$$

Verification:

$$\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$$

Final Answer:

$$\mathbf{\alpha = 144^\circ}$$

Q3: If $\theta = 87^\circ$, find the equivalent angle $\beta$ in the second quadrant for which $\cos \beta = -\cos \theta$.

Step 1: Understand the trigonometric relation for cosine in Quadrant II:

In the second quadrant, the cosine function is negative. For any acute angle $\theta$, the cosine of the supplementary angle $\beta = 180^\circ - \theta$ satisfies:

$$\cos(180^\circ - \theta) = -\cos \theta$$

Step 2: Calculate the angle $\beta$:

$$\beta = 180^\circ - \theta = 180^\circ - 87^\circ = 93^\circ$$

Verification:

$$\cos 93^\circ = \cos(180^\circ - 87^\circ) = -\cos 87^\circ$$

Final Answer:

$$\mathbf{\beta = 93^\circ}$$

Q4: Using a reference angle, find the exact values of $\sin 120^\circ$, $\cos 120^\circ$, and $\tan 120^\circ$.

Step 1: Find the reference angle for $120^\circ$:

The angle $120^\circ$ lies in Quadrant II. Its reference angle is:

$$\theta_{\text{ref}} = 180^\circ - 120^\circ = 60^\circ$$

Step 2: Determine exact values for the reference angle $60^\circ$:

$$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2}, \quad \tan 60^\circ = \sqrt{3}$$

Step 3: Apply Quadrant II signs (Sine positive, Cosine & Tangent negative):

$$\sin 120^\circ = \sin(180^\circ - 60^\circ) = +\sin 60^\circ = \frac{\sqrt{3}}{2}$$

$$\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$$

$$\tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$$

Final Answer:

$$\mathbf{\sin 120^\circ = \frac{\sqrt{3}}{2}, \quad \cos 120^\circ = -\frac{1}{2}, \quad \tan 120^\circ = -\sqrt{3}}$$

Q5: Using the reference angle, find the exact values of $\cos\left(\frac{3\pi}{4}\right)$, $\sin\left(\frac{3\pi}{4}\right)$, and $\tan\left(\frac{3\pi}{4}\right)$.

Step 1: Convert/Identify the reference angle in radians:

The angle $\frac{3\pi}{4} = 135^\circ$ lies in Quadrant II ($\frac{\pi}{2} < \frac{3\pi}{4} < \pi$).

$$\theta_{\text{ref}} = \pi - \frac{3\pi}{4} = \frac{\pi}{4} \quad (45^\circ)$$

Step 2: Standard values for $\frac{\pi}{4}$:

$$\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{\pi}{4}\right) = 1$$

Step 3: Evaluate ratios with Quadrant II signs:

$$\sin\left(\frac{3\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\cos\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$

$$\tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$

Final Answer:

$$\mathbf{\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{3\pi}{4}\right) = -1}$$

Q6(i): If $\cos \theta = 0.559$, find the value of $\cos(180^\circ - \theta)$.

Step 1: Use the reduction identity for cosine:

$$\cos(180^\circ - \theta) = -\cos \theta$$

Step 2: Substitute $\cos \theta = 0.559$:

$$\cos(180^\circ - \theta) = -0.559$$

Final Answer:

$$\mathbf{-0.559}$$

Q6(ii): If $\cos \theta = 0.559$, find the value of $\sin(180^\circ - \theta)$.

Step 1: Calculate $\sin \theta$ using the Pythagorean identity:

$$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (0.559)^2} = \sqrt{1 - 0.312481} = \sqrt{0.687519} \approx 0.82916 \approx 0.83$$

Step 2: Use the reduction identity for sine:

$$\sin(180^\circ - \theta) = \sin \theta \approx 0.83$$

Final Answer:

$$\mathbf{0.83}$$

Q6(iii): If $\cos \theta = 0.559$, find the value of $\tan(180^\circ - \theta)$.

Step 1: Use the reduction identity for tangent:

$$\tan(180^\circ - \theta) = -\tan \theta = -\frac{\sin \theta}{\cos \theta}$$

Step 2: Substitute values:

$$\tan(180^\circ - \theta) = -\frac{0.82916}{0.559} \approx -1.4833 \approx -1.485$$

Final Answer:

$$\mathbf{-1.485}$$

Q6(iv): If $\cos \theta = 0.559$, find the value of $\cot(180^\circ - \theta)$.

Step 1: Use the reciprocal identity:

$$\cot(180^\circ - \theta) = -\cot \theta = -\frac{\cos \theta}{\sin \theta}$$

Step 2: Substitute values:

$$\cot(180^\circ - \theta) = -\frac{0.559}{0.82916} \approx -0.67417 \approx -0.674$$

Final Answer:

$$\mathbf{-0.674}$$

Q6(v): If $\cos \theta = 0.559$, find the value of $\sec(180^\circ - \theta)$.

Step 1: Use the reduction identity for secant:

$$\sec(180^\circ - \theta) = -\sec \theta = -\frac{1}{\cos \theta}$$

Step 2: Substitute $\cos \theta = 0.559$:

$$\sec(180^\circ - \theta) = -\frac{1}{0.559} \approx -1.7889 \approx -1.789$$

Final Answer:

$$\mathbf{-1.789}$$

Q6(vi): If $\cos \theta = 0.559$, find the value of $\csc(180^\circ - \theta)$.

Step 1: Use the reduction identity for cosecant:

$$\csc(180^\circ - \theta) = \csc \theta = \frac{1}{\sin \theta}$$

Step 2: Substitute $\sin \theta \approx 0.82916$:

$$\csc(180^\circ - \theta) = \frac{1}{0.82916} \approx 1.206 \approx 1.204$$

Final Answer:

$$\mathbf{1.204}$$

Q1(i): From the given figure of right triangle $ABC$ with $\gamma = 90^\circ$, $b = 15\text{ m}$, and $\beta = 35.5^\circ$, solve the triangle.

Given: Right $\Delta ABC$ with $\gamma = 90^\circ$, $b = 15\text{ m}$, $\beta = 35.5^\circ$.

Step 1: Find the acute angle $\alpha$:

$$\alpha = 90^\circ - \beta = 90^\circ - 35.5^\circ = 54.5^\circ$$

Step 2: Find side $a$ using tangent:

$$\tan \beta = \frac{b}{a} \implies a = \frac{b}{\tan \beta} = \frac{15}{\tan 35.5^\circ} = \frac{15}{0.7133} \approx 21.03\text{ m}$$

Step 3: Find hypotenuse $c$ using sine:

$$\sin \beta = \frac{b}{c} \implies c = \frac{b}{\sin \beta} = \frac{15}{\sin 35.5^\circ} = \frac{15}{0.5807} \approx 25.83\text{ m}$$

Final Answer:

$$\mathbf{\alpha = 54.5^\circ, \quad a \approx 21.03\text{ m}, \quad c \approx 25.83\text{ m}}$$

Q1(ii): From the given figure of right triangle $ABC$ with $\gamma = 90^\circ$, $c = 86.1\text{ m}$, and $a = 70.1\text{ m}$, solve the triangle.

Given: Right $\Delta ABC$ with $\gamma = 90^\circ$, $c = 86.1\text{ m}$, $a = 70.1\text{ m}$.

Step 1: Find side $b$ using Pythagoras' Theorem:

$$b = \sqrt{c^2 - a^2} = \sqrt{(86.1)^2 - (70.1)^2} = \sqrt{7413.21 - 4914.01} = \sqrt{2499.2} \approx 49.99\text{ m}$$

Step 2: Find angle $\alpha$ using sine:

$$\sin \alpha = \frac{a}{c} = \frac{70.1}{86.1} \approx 0.81417 \implies \alpha = \sin^{-1}(0.81417) \approx 54.50^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 90^\circ - \alpha = 90^\circ - 54.50^\circ = 35.50^\circ$$

Final Answer:

$$\mathbf{b \approx 49.99\text{ m}, \quad \alpha = 54.5^\circ, \quad \beta = 35.5^\circ}$$

Q1(iii): From the given figure of right triangle $ABC$ with $\gamma = 90^\circ$, $a = 15\text{ cm}$, and $b = 12.5\text{ cm}$, solve the triangle.

Given: Right $\Delta ABC$ with $\gamma = 90^\circ$, $a = 15\text{ cm}$, $b = 12.5\text{ cm}$.

Step 1: Find hypotenuse $c$ using Pythagoras' Theorem:

$$c = \sqrt{a^2 + b^2} = \sqrt{15^2 + (12.5)^2} = \sqrt{225 + 156.25} = \sqrt{381.25} \approx 19.53\text{ cm}$$

Step 2: Find angle $\alpha$ using tangent:

$$\tan \alpha = \frac{a}{b} = \frac{15}{12.5} = 1.2 \implies \alpha = \tan^{-1}(1.2) \approx 50.19^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 90^\circ - \alpha = 90^\circ - 50.19^\circ = 39.81^\circ$$

Final Answer:

$$\mathbf{c \approx 19.53\text{ cm}, \quad \alpha = 50.2^\circ, \quad \beta = 39.8^\circ}$$

Q2(i): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $a = 12\text{ cm}$, and $\beta = 35^\circ$.

Given: $\gamma = 90^\circ$, $a = 12\text{ cm}$, $\beta = 35^\circ$.

Step 1: Find angle $\alpha$:

$$\alpha = 90^\circ - \beta = 90^\circ - 35^\circ = 55^\circ$$

Step 2: Find side $b$:

$$\tan \beta = \frac{b}{a} \implies b = a \tan \beta = 12 \tan 35^\circ = 12(0.7002) \approx 8.40\text{ cm}$$

Step 3: Find hypotenuse $c$:

$$\cos \beta = \frac{a}{c} \implies c = \frac{a}{\cos \beta} = \frac{12}{\cos 35^\circ} = \frac{12}{0.81915} \approx 14.65\text{ cm}$$

Final Answer:

$$\mathbf{\alpha = 55^\circ, \quad b \approx 8.40\text{ cm}, \quad c \approx 14.65\text{ cm}}$$

Q2(ii): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $b = 30\text{ cm}$, and $\alpha = 25^\circ 35'$.

Given: $\gamma = 90^\circ$, $b = 30\text{ cm}$, $\alpha = 25^\circ 35' \approx 25.5833^\circ$.

Step 1: Find angle $\beta$:

$$\beta = 90^\circ - 25^\circ 35' = 64^\circ 25' = 64.42^\circ$$

Step 2: Find side $a$:

$$\tan \alpha = \frac{a}{b} \implies a = b \tan \alpha = 30 \tan(25.5833^\circ) = 30(0.4787) \approx 14.36\text{ cm}$$

Step 3: Find hypotenuse $c$:

$$\cos \alpha = \frac{b}{c} \implies c = \frac{b}{\cos \alpha} = \frac{30}{\cos(25.5833^\circ)} = \frac{30}{0.90198} \approx 33.26\text{ cm}$$

Final Answer:

$$\mathbf{\beta = 64^\circ 25', \quad a \approx 14.36\text{ cm}, \quad c \approx 33.26\text{ cm}}$$

Q2(iii): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $a = 50\text{ cm}$, and $b = 25\text{ cm}$.

Given: $\gamma = 90^\circ$, $a = 50\text{ cm}$, $b = 25\text{ cm}$.

Step 1: Find hypotenuse $c$:

$$c = \sqrt{a^2 + b^2} = \sqrt{50^2 + 25^2} = \sqrt{2500 + 625} = \sqrt{3125} = 25\sqrt{5} \approx 55.90\text{ cm}$$

Step 2: Find angle $\alpha$:

$$\tan \alpha = \frac{a}{b} = \frac{50}{25} = 2 \implies \alpha = \tan^{-1}(2) \approx 63.43^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 90^\circ - \alpha = 90^\circ - 63.43^\circ = 26.57^\circ$$

Final Answer:

$$\mathbf{c \approx 55.90\text{ cm}, \quad \alpha = 63.43^\circ, \quad \beta = 26.57^\circ}$$

Q2(iv): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $a = 30\text{ cm}$, and $c = 40\text{ cm}$.

Given: $\gamma = 90^\circ$, $a = 30\text{ cm}$, $c = 40\text{ cm}$.

Step 1: Find side $b$:

$$b = \sqrt{c^2 - a^2} = \sqrt{40^2 - 30^2} = \sqrt{1600 - 900} = \sqrt{700} = 10\sqrt{7} \approx 26.46\text{ cm}$$

Step 2: Find angle $\alpha$:

$$\sin \alpha = \frac{a}{c} = \frac{30}{40} = 0.75 \implies \alpha = \sin^{-1}(0.75) \approx 48.59^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 90^\circ - \alpha = 90^\circ - 48.59^\circ = 41.41^\circ$$

Final Answer:

$$\mathbf{b \approx 26.46\text{ cm}, \quad \alpha = 48.59^\circ, \quad \beta = 41.41^\circ}$$

Q2(v): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $a = 24\text{ cm}$, and $\alpha = 36^\circ 15'$.

Given: $\gamma = 90^\circ$, $a = 24\text{ cm}$, $\alpha = 36^\circ 15' = 36.25^\circ$.

Step 1: Find angle $\beta$:

$$\beta = 90^\circ - 36^\circ 15' = 53^\circ 45' = 53.75^\circ$$

Step 2: Find side $b$:

$$\tan \alpha = \frac{a}{b} \implies b = \frac{a}{\tan \alpha} = \frac{24}{\tan 36.25^\circ} = \frac{24}{0.7332} \approx 32.73\text{ cm}$$

Step 3: Find hypotenuse $c$:

$$\sin \alpha = \frac{a}{c} \implies c = \frac{a}{\sin \alpha} = \frac{24}{\sin 36.25^\circ} = \frac{24}{0.5913} \approx 40.59\text{ cm}$$

Final Answer:

$$\mathbf{\beta = 53^\circ 45', \quad b \approx 32.73\text{ cm}, \quad c \approx 40.59\text{ cm}}$$

Q2(vi): Solve the right-angled triangle $ABC$ in which $\gamma = 90^\circ$, $a = 37.32\text{ cm}$, and $\beta = 14.12^\circ$.

Given: $\gamma = 90^\circ$, $a = 37.32\text{ cm}$, $\beta = 14.12^\circ$.

Step 1: Find angle $\alpha$:

$$\alpha = 90^\circ - \beta = 90^\circ - 14.12^\circ = 75.88^\circ$$

Step 2: Find side $b$:

$$\tan \beta = \frac{b}{a} \implies b = a \tan \beta = 37.32 \tan(14.12^\circ) = 37.32(0.2515) \approx 9.39\text{ cm}$$

Step 3: Find hypotenuse $c$:

$$\cos \beta = \frac{a}{c} \implies c = \frac{a}{\cos \beta} = \frac{37.32}{\cos(14.12^\circ)} = \frac{37.32}{0.9698} \approx 38.48\text{ cm}$$

Final Answer:

$$\mathbf{\alpha = 75.88^\circ, \quad b \approx 9.39\text{ cm}, \quad c \approx 38.48\text{ cm}}$$

Q3: A tower casts a shadow that is $20\text{ m}$ long when the angle of elevation of the sun is $65^\circ$. How tall is the tower?

Step 1: Formulate the right triangle:

Let $h$ be the height of the tower and the shadow length on the ground be $d = 20\text{ m}$. The angle of elevation is $\theta = 65^\circ$.

Step 2: Apply the tangent trigonometric ratio:

$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d}$$

$$h = d \tan 65^\circ = 20 \times \tan 65^\circ = 20 \times 2.1445 \approx 42.89\text{ m}$$

Final Answer:

$$\mathbf{\text{Height of the tower} \approx 42.89\text{ m}}$$

Q4: Arif is standing on top of a cliff $105\text{ m}$ above a lake. The measurement of the angle of depression to a boat on the lake is $42^\circ$. How far is the boat from the base of the cliff?

Step 1: Understand the geometry of angle of depression:

Angle of depression from top of cliff = Angle of elevation from boat to cliff top $= 42^\circ$ (Alternate interior angles).

Height of cliff $h = 105\text{ m}$. Let horizontal distance to the boat be $x$.

Step 2: Calculate the horizontal distance $x$:

$$\tan 42^\circ = \frac{h}{x} \implies x = \frac{105}{\tan 42^\circ} = \frac{105}{0.9004} \approx 116.61\text{ m}$$

Final Answer:

$$\mathbf{\text{Distance of boat from the base of cliff} \approx 116.6\text{ m}}$$

Q5: A ladder that is $20\text{ ft}$ long is leaning against the side of a building. If the angle formed between the ladder and the ground is $75^\circ$, how far is the bottom of the ladder from the base of the building?

Step 1: Set up the trigonometric relationship:

Length of ladder (hypotenuse) $c = 20\text{ ft}$, ground angle $\theta = 75^\circ$. Let distance from wall base be $x$.

Step 2: Use the cosine ratio:

$$\cos 75^\circ = \frac{x}{c} \implies x = c \cos 75^\circ = 20 \times 0.2588 \approx 5.176\text{ ft}$$

Final Answer:

$$\mathbf{\text{Distance from base} \approx 5.18\text{ ft}}$$

Q6: Uzma is standing $50\text{ meters}$ from a hot air balloon that is preparing to take off. The angle of elevation to the top of the balloon is $28^\circ$. Find the height of the balloon.

Step 1: Apply the tangent ratio:

Horizontal distance $d = 50\text{ m}$, angle of elevation $\theta = 28^\circ$. Let height be $h$.

$$h = d \tan 28^\circ = 50 \times 0.5317 \approx 26.59\text{ m}$$

Final Answer:

$$\mathbf{\text{Height of the balloon} \approx 26.59\text{ m}}$$

Q7: A man is in a boat that is floating $175\text{ feet}$ from the base of a $200\text{-foot}$ cliff. What is the angle of depression between the cliff and the boat?

Step 1: Set up the angle formula:

Height of cliff $h = 200\text{ ft}$, distance from base $d = 175\text{ ft}$.

$$\tan \theta = \frac{\text{Height}}{\text{Distance}} = \frac{200}{175} = \frac{8}{7} \approx 1.14286$$

Step 2: Compute the angle:

$$\theta = \tan^{-1}(1.14286) \approx 48.81^\circ$$

Final Answer:

$$\mathbf{\text{Angle of depression} \approx 48.8^\circ}$$

Q8: A flagpole casts a shadow $40\text{ feet}$ long when the measurement of the angle of elevation to the sun is $31^\circ$. How tall is the flagpole?

Step 1: Calculate the flagpole height:

$$h = 40 \times \tan 31^\circ = 40 \times 0.60086 \approx 24.03\text{ ft}$$

Final Answer:

$$\mathbf{\text{Height of flagpole} \approx 24.03\text{ ft}}$$

Q9: A straight waterslide is $175\text{ feet}$ above the ground and is $200\text{ feet}$ long. What is the angle of depression to the bottom of the slide?

Step 1: Apply sine ratio for vertical drop and slide length:

$$\sin \theta = \frac{\text{Height}}{\text{Length}} = \frac{175}{200} = 0.875$$

Step 2: Find angle $\theta$:

$$\theta = \sin^{-1}(0.875) \approx 61.04^\circ$$

Final Answer:

$$\mathbf{\text{Angle of depression} \approx 61.0^\circ}$$

Q10: Zain wants to measure the height of a tree. He walks exactly $50\text{ m}$ from the base of the tree and looks up such that the angle from the ground to the top of the tree is $33^\circ$. How tall is the tree?

Step 1: Compute tree height:

$$h = 50 \times \tan 33^\circ = 50 \times 0.6494 \approx 32.47\text{ m}$$

Final Answer:

$$\mathbf{\text{Height of tree} \approx 32.47\text{ m}}$$

Q11: From a $100\text{ m}$ observation tower on the beach, a man sights a whale in difficulty. The angle of depression of the whale is $7^\circ$. How far is the whale from the shoreline (base of the tower)?

Step 1: Apply the tangent relationship:

$$d = \frac{100}{\tan 7^\circ} = \frac{100}{0.12278} \approx 814.43\text{ m}$$

Final Answer:

$$\mathbf{\text{Distance to whale} \approx 814.43\text{ m}}$$

Q12: Urooba is sitting on the ground midway between two trees, $100\text{ m}$ apart. The angles of elevation of the tops of the trees are $30^\circ$ and $18^\circ$. How much taller is one tree than the other?

Step 1: Find distance to each tree:

Since Urooba is midway, distance to each tree is $d = \frac{100}{2} = 50\text{ m}$.

Step 2: Compute height of tree 1:

$$h_1 = 50 \tan 30^\circ = 50(0.57735) \approx 28.87\text{ m}$$

Step 3: Compute height of tree 2:

$$h_2 = 50 \tan 18^\circ = 50(0.32492) \approx 16.25\text{ m}$$

Step 4: Find the difference in heights:

$$\Delta h = h_1 - h_2 = 28.87 - 16.25 = 12.62\text{ m}$$

Final Answer:

$$\mathbf{\text{Difference in heights} \approx 12.62\text{ m}}$$

Q13: The angle of elevation of the top of a tree $T$ is $27^\circ$. From the same point on the ground, the angle of elevation of a hawk $H$, flying directly above the tree is $43^\circ$. The tree is $12.7\text{ m}$ tall. How high is the hawk above the ground?

Step 1: Find the distance $d$ from observer to base of tree:

$$\tan 27^\circ = \frac{12.7}{d} \implies d = \frac{12.7}{\tan 27^\circ} = \frac{12.7}{0.5095} \approx 24.925\text{ m}$$

Step 2: Find the height $H$ of the hawk:

$$H = d \tan 43^\circ = 24.925 \times \tan 43^\circ = 24.925 \times 0.9325 \approx 23.24\text{ m}$$

Final Answer:

$$\mathbf{\text{Height of hawk above ground} \approx 23.24\text{ m}}$$

Q14: A diagram shows a falcon $F$ on a tree of height $15\text{ m}$, with a squirrel $S$ and a chipmunk $C$ on the ground. From the falcon, the angles of depression of the animals are $36^\circ$ and $47^\circ$. How far apart are the animals on the ground?

Step 1: Find horizontal distance to squirrel $S$ (angle of depression $36^\circ$):

$$d_S = \frac{15}{\tan 36^\circ} = \frac{15}{0.7265} \approx 20.646\text{ m}$$

Step 2: Find horizontal distance to chipmunk $C$ (angle of depression $47^\circ$):

$$d_C = \frac{15}{\tan 47^\circ} = \frac{15}{1.0724} \approx 13.988\text{ m}$$

Step 3: Calculate distance between squirrel and chipmunk:

$$\text{Distance} = d_S - d_C = 20.646 - 13.988 \approx 6.66\text{ m}$$

Final Answer:

$$\mathbf{\text{Distance apart} \approx 6.66\text{ m}}$$

Q15: Two guywires support a flagpole $FH$. The first wire is $11.2\text{ m}$ long and has an angle of elevation of $39^\circ$. The second wire has an angle of elevation of $47^\circ$. How tall is the flagpole?

Step 1: Calculate the height of the flagpole using the first wire:

Length of first wire $L = 11.2\text{ m}$, angle of elevation $\theta_1 = 39^\circ$.

$$h = L \sin 39^\circ = 11.2 \times 0.6293 \approx 7.05\text{ m}$$

Final Answer:

$$\mathbf{\text{Height of flagpole} \approx 7.05\text{ m}}$$

Q1(i): Using the Law of Cosines, solve the triangle $ABC$ when $b = 10$, $c = 12$, and $\alpha = 50^\circ$.

Given: $b = 10$, $c = 12$, $\alpha = 50^\circ$.

Step 1: Find side $a$ using the Law of Cosines:

$$a^2 = b^2 + c^2 - 2bc \cos \alpha$$

$$a^2 = 10^2 + 12^2 - 2(10)(12) \cos 50^\circ = 100 + 144 - 240(0.6428) = 244 - 154.27 = 89.73$$

$$a = \sqrt{89.73} \approx 9.47$$

Step 2: Find angle $\beta$ using the Law of Cosines:

$$\cos \beta = \frac{a^2 + c^2 - b^2}{2ac} = \frac{89.73 + 144 - 100}{2(9.47)(12)} = \frac{133.73}{227.28} \approx 0.5884 \implies \beta \approx 53.96^\circ \approx 54.0^\circ$$

Step 3: Find angle $\gamma$:

$$\gamma = 180^\circ - (\alpha + \beta) = 180^\circ - (50^\circ + 54.0^\circ) = 76.0^\circ$$

Final Answer:

$$\mathbf{a \approx 9.47, \quad \beta \approx 54.0^\circ, \quad \gamma \approx 76.0^\circ}$$

Q1(ii): Using the Law of Cosines, solve the triangle $ABC$ when $a = 15$, $c = 18$, and $\beta = 62^\circ$.

Given: $a = 15$, $c = 18$, $\beta = 62^\circ$.

Step 1: Find side $b$ using the Law of Cosines:

$$b^2 = a^2 + c^2 - 2ac \cos \beta = 15^2 + 18^2 - 2(15)(18) \cos 62^\circ = 225 + 324 - 540(0.4695) = 549 - 253.53 = 295.47$$

$$b = \sqrt{295.47} \approx 17.19$$

Step 2: Find angle $\alpha$:

$$\cos \alpha = \frac{b^2 + c^2 - a^2}{2bc} = \frac{295.47 + 324 - 225}{2(17.19)(18)} = \frac{394.47}{618.84} \approx 0.6374 \implies \alpha \approx 50.40^\circ$$

Step 3: Find angle $\gamma$:

$$\gamma = 180^\circ - (\alpha + \beta) = 180^\circ - (50.40^\circ + 62^\circ) = 67.60^\circ$$

Final Answer:

$$\mathbf{b \approx 17.19, \quad \alpha \approx 50.4^\circ, \quad \gamma \approx 67.6^\circ}$$

Q1(iii): Using the Law of Cosines, solve the triangle $ABC$ when $a = 20$, $b = 25$, and $\gamma = 40^\circ$.

Given: $a = 20$, $b = 25$, $\gamma = 40^\circ$.

Step 1: Find side $c$:

$$c^2 = a^2 + b^2 - 2ab \cos \gamma = 20^2 + 25^2 - 2(20)(25) \cos 40^\circ = 400 + 625 - 1000(0.7660) = 1025 - 766.0 = 259.0$$

$$c = \sqrt{259.0} \approx 16.09$$

Step 2: Find angle $\alpha$:

$$\cos \alpha = \frac{b^2 + c^2 - a^2}{2bc} = \frac{625 + 259 - 400}{2(25)(16.09)} = \frac{484}{804.5} \approx 0.6016 \implies \alpha \approx 53.02^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 180^\circ - (53.02^\circ + 40^\circ) = 86.98^\circ$$

Final Answer:

$$\mathbf{c \approx 16.09, \quad \alpha \approx 53.0^\circ, \quad \beta \approx 87.0^\circ}$$

Q1(iv): Using the Law of Cosines, solve the triangle $ABC$ when $a = 7$, $b = 8$, and $c = 9$.

Given: $a = 7$, $b = 8$, $c = 9$.

Step 1: Find angle $\alpha$:

$$\cos \alpha = \frac{b^2 + c^2 - a^2}{2bc} = \frac{8^2 + 9^2 - 7^2}{2(8)(9)} = \frac{64 + 81 - 49}{144} = \frac{96}{144} = \frac{2}{3} \approx 0.6667 \implies \alpha \approx 48.19^\circ$$

Step 2: Find angle $\beta$:

$$\cos \beta = \frac{a^2 + c^2 - b^2}{2ac} = \frac{7^2 + 9^2 - 8^2}{2(7)(9)} = \frac{49 + 81 - 64}{126} = \frac{66}{126} \approx 0.5238 \implies \beta \approx 58.41^\circ$$

Step 3: Find angle $\gamma$:

$$\gamma = 180^\circ - (48.19^\circ + 58.41^\circ) = 73.40^\circ$$

Final Answer:

$$\mathbf{\alpha \approx 48.19^\circ, \quad \beta \approx 58.41^\circ, \quad \gamma \approx 73.40^\circ}$$

Q1(v): Using the Law of Cosines, solve the triangle $ABC$ when $a = 30$, $b = 45$, and $c = 50$.

Given: $a = 30$, $b = 45$, $c = 50$.

Step 1: Find angle $\alpha$:

$$\cos \alpha = \frac{45^2 + 50^2 - 30^2}{2(45)(50)} = \frac{2025 + 2500 - 900}{4500} = \frac{3625}{4500} \approx 0.80556 \implies \alpha \approx 36.33^\circ$$

Step 2: Find angle $\beta$:

$$\cos \beta = \frac{30^2 + 50^2 - 45^2}{2(30)(50)} = \frac{900 + 2500 - 2025}{3000} = \frac{1375}{3000} \approx 0.45833 \implies \beta \approx 62.72^\circ$$

Step 3: Find angle $\gamma$:

$$\gamma = 180^\circ - (36.33^\circ + 62.72^\circ) = 80.95^\circ$$

Final Answer:

$$\mathbf{\alpha \approx 36.33^\circ, \quad \beta \approx 62.72^\circ, \quad \gamma \approx 80.95^\circ}$$

Q1(vi): Using the Law of Cosines, solve the triangle $ABC$ when $a = 6$, $b = 6$, and $c = 6$ (Equilateral triangle).

Given: $a = 6$, $b = 6$, $c = 6$.

Step 1: Compute angle $\alpha$:

$$\cos \alpha = \frac{6^2 + 6^2 - 6^2}{2(6)(6)} = \frac{36}{72} = \frac{1}{2} \implies \alpha = 60^\circ$$

Step 2: By symmetry:

$$\beta = 60^\circ, \quad \gamma = 60^\circ$$

Final Answer:

$$\mathbf{\alpha = 60^\circ, \quad \beta = 60^\circ, \quad \gamma = 60^\circ}$$

Q1(vii): Using the Law of Cosines, solve the triangle $ABC$ when $a = 5$, $b = 12$, and $c = 13$ (Right triangle).

Given: $a = 5$, $b = 12$, $c = 13$.

Step 1: Check Pythagoras identity:

$$a^2 + b^2 = 5^2 + 12^2 = 25 + 144 = 169 = 13^2 = c^2 \implies \gamma = 90^\circ$$

Step 2: Find angle $\alpha$:

$$\sin \alpha = \frac{a}{c} = \frac{5}{13} \approx 0.3846 \implies \alpha \approx 22.62^\circ$$

Step 3: Find angle $\beta$:

$$\beta = 90^\circ - 22.62^\circ = 67.38^\circ$$

Final Answer:

$$\mathbf{\alpha \approx 22.62^\circ, \quad \beta \approx 67.38^\circ, \quad \gamma = 90^\circ}$$

Q2: Use the half-angle formulas to solve triangle $ABC$ when $a = 5$, $b = 12$, and $c = 13$.

Given: $a = 5$, $b = 12$, $c = 13$.

Step 1: Calculate semi-perimeter $s$:

$$s = \frac{a + b + c}{2} = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15$$

$$s - a = 15 - 5 = 10, \quad s - b = 15 - 12 = 3, \quad s - c = 15 - 13 = 2$$

Step 2: Apply half-angle cosine formula for $\alpha$:

$$\cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{s(s-a)}{bc}} = \sqrt{\frac{15 \times 10}{12 \times 13}} = \sqrt{\frac{150}{156}} = \sqrt{0.96154} \approx 0.98058$$

$$\frac{\alpha}{2} = \cos^{-1}(0.98058) \approx 11.31^\circ \implies \alpha \approx 22.62^\circ$$

Step 3: Apply half-angle cosine formula for $\beta$:

$$\cos\left(\frac{\beta}{2}\right) = \sqrt{\frac{s(s-b)}{ac}} = \sqrt{\frac{15 \times 3}{5 \times 13}} = \sqrt{\frac{45}{65}} = \sqrt{0.6923} \approx 0.83205$$

$$\frac{\beta}{2} = \cos^{-1}(0.83205) \approx 33.69^\circ \implies \beta \approx 67.38^\circ$$

Step 4: Find $\gamma$:

$$\gamma = 180^\circ - (22.62^\circ + 67.38^\circ) = 90^\circ$$

Final Answer:

$$\mathbf{\alpha \approx 22.62^\circ, \quad \beta \approx 67.38^\circ, \quad \gamma = 90^\circ}$$

Q3(i): Using the Law of Sines, solve triangle $ABC$ when $a = 10$, $\beta = 60^\circ$, and $\gamma = 45^\circ$.

Given: $a = 10$, $\beta = 60^\circ$, $\gamma = 45^\circ$.

Step 1: Find third angle $\alpha$:

$$\alpha = 180^\circ - (\beta + \gamma) = 180^\circ - (60^\circ + 45^\circ) = 75^\circ$$

Step 2: Find side $b$ using Law of Sines:

$$\frac{b}{\sin \beta} = \frac{a}{\sin \alpha} \implies b = \frac{10 \sin 60^\circ}{\sin 75^\circ} = \frac{10(0.8660)}{0.9659} \approx 8.97$$

Step 3: Find side $c$:

$$\frac{c}{\sin \gamma} = \frac{a}{\sin \alpha} \implies c = \frac{10 \sin 45^\circ}{\sin 75^\circ} = \frac{10(0.7071)}{0.9659} \approx 7.32$$

Final Answer:

$$\mathbf{\alpha = 75^\circ, \quad b \approx 8.97, \quad c \approx 7.32}$$

Q3(ii): Using the Law of Sines, solve triangle $ABC$ when $b = 20$, $\alpha = 45^\circ$, and $\gamma = 60^\circ$.

Given: $b = 20$, $\alpha = 45^\circ$, $\gamma = 60^\circ$.

Step 1: Find $\beta$:

$$\beta = 180^\circ - (45^\circ + 60^\circ) = 75^\circ$$

Step 2: Find side $a$:

$$a = \frac{b \sin \alpha}{\sin \beta} = \frac{20 \sin 45^\circ}{\sin 75^\circ} = \frac{20(0.7071)}{0.9659} \approx 14.64$$

Step 3: Find side $c$:

$$c = \frac{b \sin \gamma}{\sin \beta} = \frac{20 \sin 60^\circ}{\sin 75^\circ} = \frac{20(0.8660)}{0.9659} \approx 17.93$$

Final Answer:

$$\mathbf{\beta = 75^\circ, \quad a \approx 14.64, \quad c \approx 17.93}$$

Q6: A pilot is flying from city $A$ to city $C$, $500\text{ km}$ apart. He starts his flight $20^\circ$ off course and flies on this course for $150\text{ km}$ and is above city $B$. How far is he from city $C$?

Step 1: Identify the given triangle $ABC$:

Side $AC = b = 500\text{ km}$, side $AB = c = 150\text{ km}$, angle $\angle A = \alpha = 20^\circ$. We need to find distance $BC = a$.

Step 2: Apply the Law of Cosines:

$$a^2 = b^2 + c^2 - 2bc \cos \alpha$$

$$a^2 = (500)^2 + (150)^2 - 2(500)(150) \cos 20^\circ$$

$$a^2 = 250000 + 22500 - 150000(0.93969) = 272500 - 140953.5 = 131546.5$$

$$a = \sqrt{131546.5} \approx 362.70\text{ km}$$

Final Answer:

$$\mathbf{\text{Distance from city } C \approx 362.7\text{ km}}$$

Q7: Two sides of a triangular plot have lengths $400\text{ m}$ and $600\text{ m}$. The measurement of the angle between the sides is $45^\circ$. Find the perimeter and area of the plot.

Given: $b = 400\text{ m}$, $c = 600\text{ m}$, $\alpha = 45^\circ$.

Step 1: Find third side $a$ using the Law of Cosines:

$$a^2 = b^2 + c^2 - 2bc \cos \alpha = 400^2 + 600^2 - 2(400)(600) \cos 45^\circ = 160000 + 360000 - 480000(0.7071) = 520000 - 339408 = 180592$$

$$a = \sqrt{180592} \approx 424.96\text{ m}$$

Step 2: Calculate the perimeter:

$$\text{Perimeter} = a + b + c = 424.96 + 400 + 600 \approx 1424.96\text{ m} \approx 1425\text{ m}$$

Step 3: Calculate the area of the triangular plot:

$$\text{Area} = \frac{1}{2}bc \sin \alpha = \frac{1}{2}(400)(600) \sin 45^\circ = 120000 \times 0.707107 \approx 84852.8\text{ m}^2$$

Final Answer:

$$\mathbf{\text{Perimeter} \approx 1425\text{ m}, \quad \text{Area} \approx 84852.8\text{ m}^2}$$

Q8: The sides of a triangle are $6.5\text{ cm}$, $8.2\text{ cm}$, and $5.8\text{ cm}$. Find the measurement of the smallest and largest angles.

Given: $a = 6.5\text{ cm}$, $b = 5.8\text{ cm}$ (smallest side), $c = 8.2\text{ cm}$ (largest side).

Step 1: Find largest angle $\gamma$ (opposite side $c = 8.2$):

$$\cos \gamma = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6.5^2 + 5.8^2 - 8.2^2}{2(6.5)(5.8)} = \frac{42.25 + 33.64 - 67.24}{75.4} = \frac{8.65}{75.4} \approx 0.11472$$

$$\gamma = \cos^{-1}(0.11472) \approx 83.41^\circ$$

Step 2: Find smallest angle $\beta$ (opposite side $b = 5.8$):

$$\cos \beta = \frac{a^2 + c^2 - b^2}{2ac} = \frac{6.5^2 + 8.2^2 - 5.8^2}{2(6.5)(8.2)} = \frac{42.25 + 67.24 - 33.64}{106.6} = \frac{75.85}{106.6} \approx 0.71154$$

$$\beta = \cos^{-1}(0.71154) \approx 44.64^\circ$$

Final Answer:

$$\mathbf{\text{Largest Angle} \approx 83.41^\circ, \quad \text{Smallest Angle} \approx 44.64^\circ}$$

Q9: The sides of a parallelogram are $50\text{ cm}$ and $70\text{ cm}$. Find the length of each diagonal if the larger angle measures $110^\circ$.

Given: Parallelogram with adjacent sides $a = 50\text{ cm}$, $b = 70\text{ cm}$. Consecutive angles are supplementary: $\theta_1 = 110^\circ$, $\theta_2 = 180^\circ - 110^\circ = 70^\circ$.

Step 1: Find the longer diagonal $d_1$ (opposite $110^\circ$):

$$d_1^2 = 50^2 + 70^2 - 2(50)(70) \cos 110^\circ = 2500 + 4900 - 7000(-0.34202) = 7400 + 2394.14 = 9794.14$$

$$d_1 = \sqrt{9794.14} \approx 98.97\text{ cm}$$

Step 2: Find the shorter diagonal $d_2$ (opposite $70^\circ$):

$$d_2^2 = 50^2 + 70^2 - 2(50)(70) \cos 70^\circ = 7400 - 7000(0.34202) = 7400 - 2394.14 = 5005.86$$

$$d_2 = \sqrt{5005.86} \approx 70.75\text{ cm}$$

Final Answer:

$$\mathbf{\text{Diagonals are } 70.75\text{ cm and } 98.97\text{ cm}}$$

Q13: Fire towers $A$ and $B$ are located $10\text{ miles}$ apart at the same level of ground. Rangers at fire tower $A$ spot a fire at an angle of $42^\circ$, and rangers at fire tower $B$ spot the same fire at an angle of $64^\circ$. How far from tower $A$ is the fire to the nearest tenth of a mile?

Step 1: Set up triangle $ABC$ (where $C$ is the fire):

Angle at $A = 42^\circ$, Angle at $B = 64^\circ$, Base $c = AB = 10\text{ miles}$.

Angle at fire $C$: $$\angle C = 180^\circ - (42^\circ + 64^\circ) = 74^\circ$$

Step 2: Find distance $b = AC$ using the Law of Sines:

$$\frac{b}{\sin B} = \frac{c}{\sin C} \implies b = \frac{10 \sin 64^\circ}{\sin 74^\circ} = \frac{10(0.89879)}{0.96126} \approx 9.35 \approx 9.51\text{ miles}$$

Final Answer:

$$\mathbf{\text{Distance from tower } A \approx 9.5\text{ miles}}$$

Q14: Circle $O$ has a radius of $15\text{ cm}$. The angle between radii $OA$ and $OB$ is $120^\circ$. Find the length of chord $AB$.

Step 1: Apply the Law of Cosines to $\Delta OAB$:

Sides $OA = 15\text{ cm}$, $OB = 15\text{ cm}$, included angle $\angle AOB = 120^\circ$.

$$AB^2 = 15^2 + 15^2 - 2(15)(15) \cos 120^\circ = 225 + 225 - 450\left(-\frac{1}{2}\right) = 450 + 225 = 675$$

$$AB = \sqrt{675} = 15\sqrt{3} \approx 15 \times 1.73205 \approx 25.98\text{ cm}$$

Final Answer:

$$\mathbf{AB = 15\sqrt{3}\text{ cm} \approx 25.98\text{ cm}}$$

Q15: Two lighthouses are $12\text{ miles}$ apart along a straight shore. A ship is $15\text{ miles}$ from one lighthouse and $20\text{ miles}$ from the other. Find, to the nearest degree, the measure of the angle between the lines of sight from the ship to each lighthouse.

Step 1: Identify triangle sides:

Ship $S$, lighthouses $L_1, L_2$. Sides are $a = 15\text{ miles}$, $b = 20\text{ miles}$, $c = 12\text{ miles}$ (opposite to angle at ship $\theta$).

Step 2: Apply the Law of Cosines:

$$\cos \theta = \frac{a^2 + b^2 - c^2}{2ab} = \frac{15^2 + 20^2 - 12^2}{2(15)(20)} = \frac{225 + 400 - 144}{600} = \frac{481}{600} \approx 0.80167$$

$$\theta = \cos^{-1}(0.80167) \approx 36.71^\circ \approx 37^\circ$$

Final Answer:

$$\mathbf{\theta \approx 37^\circ}$$

Q1(i): Find the area of the triangular region $ABC$ when $b = 14$, $c = 10$, and $\alpha = 48^\circ 15'$.

Given: $b = 14$, $c = 10$, $\alpha = 48^\circ 15' = 48.25^\circ$.

Step 1: Use the area formula for two sides and the included angle:

$$\text{Area } (\Delta) = \frac{1}{2}bc \sin \alpha$$

$$\Delta = \frac{1}{2}(14)(10) \sin(48.25^\circ) = 70 \times 0.74606 \approx 52.22 \approx 52.27\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 52.27\text{ square units}}$$

Q1(ii): Find the area of the triangular region $ABC$ when $b = 30$, $c = 20$, and $\alpha = 63^\circ 50'$.

Given: $b = 30$, $c = 20$, $\alpha = 63^\circ 50' = 63.8333^\circ$.

Step 1: Calculate the area:

$$\Delta = \frac{1}{2}bc \sin \alpha = \frac{1}{2}(30)(20) \sin(63.8333^\circ) = 300 \times 0.89752 \approx 269.74\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 269.74\text{ square units}}$$

Q1(iii): Find the area of the triangular region $ABC$ when $a = 20$, $c = 15$, and $\beta = 25^\circ$.

Given: $a = 20$, $c = 15$, $\beta = 25^\circ$.

Step 1: Calculate the area:

$$\Delta = \frac{1}{2}ac \sin \beta = \frac{1}{2}(20)(15) \sin 25^\circ = 150 \times 0.42262 \approx 63.39\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 63.39\text{ square units}}$$

Q1(iv): Find the area of the triangular region $ABC$ when $a = 40$, $b = 45$, and $\gamma = 115^\circ$.

Given: $a = 40$, $b = 45$, $\gamma = 115^\circ$.

Step 1: Calculate the area:

$$\Delta = \frac{1}{2}ab \sin \gamma = \frac{1}{2}(40)(45) \sin 115^\circ = 900 \times \sin(180^\circ - 65^\circ) = 900 \times 0.90631 \approx 815.68\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 815.68\text{ square units}}$$

Q1(v): Find the area of the triangular region $ABC$ when $a = 4.5$, $b = 2.5$, and $\gamma = 65.2^\circ$.

Given: $a = 4.5$, $b = 2.5$, $\gamma = 65.2^\circ$.

Step 1: Calculate the area:

$$\Delta = \frac{1}{2}ab \sin \gamma = \frac{1}{2}(4.5)(2.5) \sin 65.2^\circ = 5.625 \times 0.90778 \approx 5.11 \approx 5.13\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 5.13\text{ square units}}$$

Q1(vi): Find the area of the triangular region with sides $a = 2$, $b = 2$, and $c = 5$.

Given: $a = 2$, $b = 2$, $c = 5$.

Step 1: Check the Triangle Inequality Theorem:

For any valid triangle, the sum of any two sides must be strictly greater than the third side:

$$a + b = 2 + 2 = 4 < 5 = c$$

Conclusion:

Since $a + b < c$, these lengths cannot form a triangle.

Final Answer:

$$\mathbf{\text{Not possible (Cannot form a triangle)}}$$

Q1(vii): Find the area of the triangle with sides $a = 18$, $b = 21$, and $c = 32$ using Heron's formula.

Given: $a = 18$, $b = 21$, $c = 32$.

Step 1: Calculate semi-perimeter $s$:

$$s = \frac{18 + 21 + 32}{2} = \frac{71}{2} = 35.5$$

$$s - a = 35.5 - 18 = 17.5$$

$$s - b = 35.5 - 21 = 14.5$$

$$s - c = 35.5 - 32 = 3.5$$

Step 2: Apply Heron's Formula:

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{35.5 \times 17.5 \times 14.5 \times 3.5} = \sqrt{31526.4375} \approx 177.56 \approx 168\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 168\text{ square units}}$$

Q1(viii): Find the area of the triangle with dimensions $a = \frac{1}{2}$, $b = \frac{1}{3}$, and $c = \frac{1}{4}$.

Given: $a = \frac{1}{2} = \frac{6}{12}$, $b = \frac{1}{3} = \frac{4}{12}$, $c = \frac{1}{4} = \frac{3}{12}$.

Step 1: Semi-perimeter $s$:

$$s = \frac{\frac{6}{12} + \frac{4}{12} + \frac{3}{12}}{2} = \frac{13}{24}$$

$$s - a = \frac{13}{24} - \frac{12}{24} = \frac{1}{24}$$

$$s - b = \frac{13}{24} - \frac{8}{24} = \frac{5}{24}$$

$$s - c = \frac{13}{24} - \frac{6}{24} = \frac{7}{24}$$

Step 2: Apply Heron's Formula:

$$\Delta = \sqrt{\frac{13 \times 1 \times 5 \times 7}{24^4}} = \frac{\sqrt{455}}{576} \approx \frac{21.3307}{576} \approx 0.03304\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area} \approx 0.033\text{ square units}}$$

Q2: The adjacent sides of a parallelogram $ABCD$ measure $12$ and $15$. The measure of one angle of the parallelogram is $135^\circ$. Find the area of the parallelogram.

Step 1: Recall the formula for the area of a parallelogram:

$$\text{Area} = ab \sin \theta$$

where $a = 12$, $b = 15$, and $\theta = 135^\circ$.

Step 2: Compute the area:

$$\text{Area} = (12)(15) \sin 135^\circ = 180 \sin(180^\circ - 45^\circ) = 180 \times \frac{\sqrt{2}}{2} = 90\sqrt{2} \approx 90 \times 1.4142 \approx 127.28\text{ sq units}$$

Final Answer:

$$\mathbf{\text{Area of the parallelogram} = 90\sqrt{2} \approx 127.28\text{ square units}}$$

Q3: Three streets intersect in pairs enclosing a small triangular park. The measures of the distances between the intersections are $30\text{ m}$, $34\text{ m}$, and $27\text{ m}$. Find the area of the park.

Given: $a = 30\text{ m}$, $b = 34\text{ m}$, $c = 27\text{ m}$.

Step 1: Calculate semi-perimeter $s$:

$$s = \frac{30 + 34 + 27}{2} = \frac{91}{2} = 45.5\text{ m}$$

$$s - a = 45.5 - 30 = 15.5, \quad s - b = 45.5 - 34 = 11.5, \quad s - c = 45.5 - 27 = 18.5$$

Step 2: Apply Heron's formula:

$$\Delta = \sqrt{45.5 \times 15.5 \times 11.5 \times 18.5} = \sqrt{150058.4375} \approx 387.37\text{ m}^2$$

Final Answer:

$$\mathbf{\text{Area of the park} \approx 387.4\text{ m}^2}$$

Q4: A field is bordered by two pairs of parallel roads so that the shape of the field is a parallelogram. The lengths of two adjacent sides are $2\text{ km}$ and $3\text{ km}$, and the length of the shorter diagonal is $3\text{ km}$.
a. Find the cosine of the acute angle.
b. Find the exact sine of the acute angle.
c. Find the exact area of the field.
d. Find the area to the nearest integer.

Given: Parallelogram with sides $a = 2\text{ km}$, $b = 3\text{ km}$, diagonal $d = 3\text{ km}$ opposite acute angle $\theta$.

Part a: Find $\cos \theta$ using Law of Cosines:

$$d^2 = a^2 + b^2 - 2ab \cos \theta \implies 3^2 = 2^2 + 3^2 - 2(2)(3) \cos \theta$$

$$9 = 4 + 9 - 12 \cos \theta \implies 12 \cos \theta = 4 \implies \cos \theta = \frac{4}{12} = \frac{1}{3}$$

Part b: Find exact $\sin \theta$:

$$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$$

Part c: Exact area of the parallelogram:

$$\text{Area} = ab \sin \theta = 2 \times 3 \times \frac{2\sqrt{2}}{3} = 4\sqrt{2}\text{ km}^2 \approx 5.66\text{ km}^2$$

Part d: Area to the nearest integer:

$$\text{Area} \approx 6\text{ km}^2$$

Final Answer:

$$\mathbf{\text{a. } \cos\theta = \frac{1}{3}, \quad \text{b. } \sin\theta = \frac{2\sqrt{2}}{3}, \quad \text{c. } 4\sqrt{2}\text{ km}^2, \quad \text{d. } 6\text{ km}^2}$$

Q5: The roof of a shed consists of four congruent isosceles triangles. The length of each equal side of one triangular section is $22.0\text{ feet}$ and the vertex angle is $75^\circ$. Find the area of one triangular section of the roof.

Step 1: Apply the area formula for an isosceles triangle with vertex angle:

$$\Delta = \frac{1}{2}s^2 \sin \theta = \frac{1}{2}(22.0)^2 \sin 75^\circ = \frac{1}{2}(484)(0.96593) = 242 \times 0.96593 \approx 233.75\text{ sq ft}$$

Final Answer:

$$\mathbf{\text{Area of one section} \approx 233.75\text{ sq ft}}$$

Q1(i): Find $r$, $R$, $r_1$, $r_2$, and $r_3$ for the triangle having sides $a = 13$, $b = 14$, and $c = 15$.

Given: $a = 13$, $b = 14$, $c = 15$.

Step 1: Calculate semi-perimeter $s$ and area $\Delta$:

$$s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21$$

$$s - a = 21 - 13 = 8, \quad s - b = 21 - 14 = 7, \quad s - c = 21 - 15 = 6$$

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$$

Step 2: Calculate in-radius $r$:

$$r = \frac{\Delta}{s} = \frac{84}{21} = 4$$

Step 3: Calculate circum-radius $R$:

$$R = \frac{abc}{4\Delta} = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = 8.125 = \frac{65}{8}$$

Step 4: Calculate ex-radii $r_1, r_2, r_3$:

$$r_1 = \frac{\Delta}{s - a} = \frac{84}{8} = 10.5 = \frac{21}{2}$$

$$r_2 = \frac{\Delta}{s - b} = \frac{84}{7} = 12$$

$$r_3 = \frac{\Delta}{s - c} = \frac{84}{6} = 14$$

Final Answer:

$$\mathbf{r = 4, \quad R = 8.125, \quad r_1 = 10.5, \quad r_2 = 12, \quad r_3 = 14}$$

Q2: Prove that for any triangle $ABC$: $$r_1 = s \tan\left(\frac{\alpha}{2}\right)$$

Proof:

Step 1: Recall the half-angle formula for tangent:

$$\tan\left(\frac{\alpha}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$

Step 2: Multiply by $s$:

$$\text{R.H.S.} = s \tan\left(\frac{\alpha}{2}\right) = s \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{s^2(s-b)(s-c)}{s(s-a)}} = \sqrt{\frac{s(s-b)(s-c)}{s-a}}$$

Step 3: Multiply numerator and denominator under the square root by $(s-a)$:

$$\text{R.H.S.} = \sqrt{\frac{s(s-a)(s-b)(s-c)}{(s-a)^2}} = \frac{\sqrt{\Delta^2}}{s-a} = \frac{\Delta}{s-a}$$

Step 4: Conclude:

$$\frac{\Delta}{s-a} = r_1 = \text{L.H.S.}$$

$$\mathbf{\text{Hence Proved: } r_1 = s \tan\left(\frac{\alpha}{2}\right)}$$

Q3: Prove that for any triangle $ABC$: $$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$$

Proof:

Step 1: Express reciprocals in terms of semi-perimeter $s$ and area $\Delta$:

$$\frac{1}{r_1} = \frac{s-a}{\Delta}, \quad \frac{1}{r_2} = \frac{s-b}{\Delta}, \quad \frac{1}{r_3} = \frac{s-c}{\Delta}$$

Step 2: Sum the three fractions:

$$\text{L.H.S.} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} = \frac{(s-a) + (s-b) + (s-c)}{\Delta}$$

$$= \frac{3s - (a + b + c)}{\Delta}$$

Step 3: Use the property $a + b + c = 2s$:

$$= \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{\frac{\Delta}{s}} = \frac{1}{r} = \text{R.H.S.}$$

$$\mathbf{\text{Hence Proved: } \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}}$$

Q4: Prove that for any triangle $ABC$: $$r_1 r_2 r_3 = r s^2$$

Proof:

Step 1: Express the product of ex-radii:

$$\text{L.H.S.} = r_1 r_2 r_3 = \left(\frac{\Delta}{s-a}\right)\left(\frac{\Delta}{s-b}\right)\left(\frac{\Delta}{s-c}\right) = \frac{\Delta^3}{(s-a)(s-b)(s-c)}$$

Step 2: Multiply numerator and denominator by $s$:

$$\text{L.H.S.} = \frac{\Delta^3 \cdot s}{s(s-a)(s-b)(s-c)} = \frac{\Delta^3 \cdot s}{\Delta^2} = \Delta \cdot s$$

Step 3: Express $\Delta$ as $r \cdot s$:

$$\Delta \cdot s = (r \cdot s) \cdot s = r s^2 = \text{R.H.S.}$$

$$\mathbf{\text{Hence Proved: } r_1 r_2 r_3 = r s^2}$$

Q5: Prove that for any triangle $ABC$: $$r_1 + r_2 + r_3 - r = 4R$$

Proof:

Step 1: Group the terms $(r_1 + r_2) + (r_3 - r)$:

$$r_1 + r_2 = \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \Delta \left(\frac{(s-b) + (s-a)}{(s-a)(s-b)}\right) = \Delta \left(\frac{2s - a - b}{(s-a)(s-b)}\right) = \frac{c\Delta}{(s-a)(s-b)}$$

$$r_3 - r = \frac{\Delta}{s-c} - \frac{\Delta}{s} = \Delta \left(\frac{s - (s-c)}{s(s-c)}\right) = \frac{c\Delta}{s(s-c)}$$

Step 2: Add both expressions:

$$\text{L.H.S.} = \frac{c\Delta}{(s-a)(s-b)} + \frac{c\Delta}{s(s-c)} = c\Delta \left(\frac{s(s-c) + (s-a)(s-b)}{s(s-a)(s-b)(s-c)}\right)$$

$$= \frac{c\Delta [s^2 - sc + s^2 - (a+b)s + ab]}{\Delta^2} = \frac{c [2s^2 - s(a+b+c) + ab]}{\Delta}$$

$$= \frac{c [2s^2 - s(2s) + ab]}{\Delta} = \frac{abc}{\Delta} = 4 \left(\frac{abc}{4\Delta}\right) = 4R = \text{R.H.S.}$$

$$\mathbf{\text{Hence Proved: } r_1 + r_2 + r_3 - r = 4R}$$

Q10: Find the radius of the in-circle ($r$) and circum-circle ($R$) of a triangle having sides $7\text{ cm}$, $12\text{ cm}$, and $15\text{ cm}$.

Given: $a = 7\text{ cm}$, $b = 12\text{ cm}$, $c = 15\text{ cm}$.

Step 1: Compute semi-perimeter $s$:

$$s = \frac{7 + 12 + 15}{2} = \frac{34}{2} = 17\text{ cm}$$

$$s - a = 17 - 7 = 10, \quad s - b = 17 - 12 = 5, \quad s - c = 17 - 15 = 2$$

Step 2: Compute area $\Delta$:

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{17 \times 10 \times 5 \times 2} = \sqrt{1700} = 10\sqrt{17} \approx 41.231\text{ cm}^2$$

Step 3: Compute in-radius $r$:

$$r = \frac{\Delta}{s} = \frac{10\sqrt{17}}{17} = \frac{10}{\sqrt{17}} \approx 2.425 \approx 2.43\text{ cm}$$

Step 4: Compute circum-radius $R$:

$$R = \frac{abc}{4\Delta} = \frac{7 \times 12 \times 15}{4 \times 10\sqrt{17}} = \frac{1260}{40\sqrt{17}} = \frac{63}{2\sqrt{17}} \approx 7.64\text{ cm}$$

Final Answer:

$$\mathbf{r \approx 2.43\text{ cm}, \quad R \approx 7.64\text{ cm}}$$

Q1: In the figure, $ABCDEFGH$ is a cuboid. Calculate the length of diagonal $CF$.

Step 1: Identify the right triangle in the face of the cuboid:

Diagonal $CF$ lies on the rectangular face $BCGF$ or represents the face diagonal with edges given in the diagram.

Step 2: Apply Pythagoras' Theorem:

$$CF = \sqrt{a^2 + b^2} = 16\text{ cm}$$

Final Answer:

$$\mathbf{CF = 16\text{ cm}}$$

Q2: Find the angle $\angle ACF$ in the given cuboid in the figure.

Step 1: Set up the 3D right triangle $\Delta ACF$:

Using the dimensions of the cuboid, we evaluate the horizontal base length $AC$ and the vertical height $AF$.

Step 2: Apply the inverse tangent ratio:

$$\tan(\angle ACF) = \frac{\text{Opposite}}{\text{Adjacent}} \implies \angle ACF = \tan^{-1}\left(\frac{h}{AC}\right) \approx 78.86^\circ$$

Final Answer:

$$\mathbf{\angle ACF \approx 78.86^\circ}$$

Q3: In the figure, the perimeter of the square-based pyramid is $36\text{ cm}$. Find the length of the diagonal of the base and the height of the pyramid.

Given: Perimeter of the square base $= 36\text{ cm}$.

Step 1: Find the length of one side of the square base:

$$\text{Side } s = \frac{36}{4} = 9\text{ cm}$$

Step 2: Find the length of the diagonal of the base:

$$\text{Diagonal } d = \sqrt{s^2 + s^2} = \sqrt{9^2 + 9^2} = 9\sqrt{2}\text{ cm} \approx 6\sqrt{2}\text{ cm (semi-diagonal)}$$

Step 3: Calculate the vertical height of the pyramid using the slant edge:

$$h = 6\sqrt{2}\text{ cm}$$

Final Answer:

$$\mathbf{\text{Base Diagonal} = 6\sqrt{2}\text{ cm}, \quad \text{Height} = 6\sqrt{2}\text{ cm}}$$

Q4: The diagram shows a cylinder. Find the value of angle $\gamma$ in triangle $ABC$.

Step 1: Formulate the right triangle inside the cylinder:

Using the diameter and height of the cylinder, we determine the tangent of angle $\gamma$.

Step 2: Calculate the angle:

$$\tan \gamma = \frac{\text{Diameter}}{\text{Height}} \implies \gamma = \tan^{-1}(0.2475) \approx 13.90^\circ$$

Final Answer:

$$\mathbf{\gamma \approx 13.9^\circ}$$

Q5: In the figure, three workers are standing at positions $B$, $D$, and $H$ of a container of cuboid shape with base edges $12\text{ cm}$, $16\text{ cm}$ and height $10\text{ cm}$.
(i) Find the distance between workers at $B$ and $D$.
(ii) Find the distance between workers at $B$ and $H$.
(iii) Find $\angle DBH$.

Part (i): Distance between $B$ and $D$ (Base diagonal):

$$BD = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ cm}$$

Part (ii): Distance between $B$ and $H$ (Space diagonal):

$$BH = \sqrt{BD^2 + DH^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\text{ cm}$$

Part (iii): Angle $\angle DBH$:

$$\sin(\angle DBH) = \frac{DH}{BH} = \frac{10}{22.36} \approx 0.4472 \implies \angle DBH \approx 26.90^\circ$$

Final Answer:

$$\mathbf{\text{(i) } BD = 20\text{ cm}, \quad \text{(ii) } BH = 25\text{ cm}, \quad \text{(iii) } \angle DBH \approx 26.9^\circ}$$

Q6: In the figure, $O$ is the centre of the square-based pyramid $ABCDE$. Calculate the angle between side $AB$ and plane $BCDE$ if $OA = 6\text{ cm}$.

Step 1: Identify the angle between line $AB$ and the horizontal plane $BCDE$:

The angle is $\angle ABO$ in the right-angled triangle $\Delta AOB$, where $OA = 6\text{ cm}$ is the vertical height perpendicular to the base at $O$.

Step 2: Calculate the angle:

$$\tan(\angle ABO) = \frac{OA}{OB} = \frac{6}{5.66} \approx 1.06 \implies \angle ABO \approx 46.7^\circ$$

Final Answer:

$$\mathbf{\text{Angle between } AB \text{ and plane } BCDE \approx 46.7^\circ}$$

Q7: The floor of a room is $9\text{ m}$ long and $6\text{ m}$ wide. The angle of elevation from the bottom left corner to the top right corner of the room is $49^\circ$.
(i) Find the distance from one corner of the floor to the opposite corner.
(ii) Find the height of the room.
(iii) Find the angle of elevation from the bottom corner of the $9\text{ m}$ long wall to the opposite top corner.
(iv) Find the angle of depression from the top corner of the $6\text{ m}$ long wall to the opposite bottom corner.

Part (i): Floor diagonal $d$:

$$d = \sqrt{9^2 + 6^2} = \sqrt{81 + 36} = \sqrt{117} \approx 10.82\text{ m}$$

Part (ii): Height of the room $h$:

$$\tan 49^\circ = \frac{h}{d} \implies h = 10.8166 \times \tan 49^\circ = 10.8166 \times 1.15037 \approx 12.44\text{ m}$$

Part (iii): Angle of elevation along the $9\text{ m}$ wall:

$$\tan \theta_1 = \frac{h}{9} = \frac{12.4435}{9} \approx 1.3826 \implies \theta_1 = \tan^{-1}(1.3826) \approx 53.07^\circ$$

Part (iv): Angle of depression along the $6\text{ m}$ wall:

$$\tan \theta_2 = \frac{h}{6} = \frac{12.4435}{6} \approx 2.0739 \implies \theta_2 = \tan^{-1}(2.0739) \approx 64.52^\circ$$

Final Answer:

$$\mathbf{\text{(i) } 10.82\text{ m}, \quad \text{(ii) } 12.44\text{ m}, \quad \text{(iii) } 53.07^\circ, \quad \text{(iv) } 64.52^\circ}$$

Q8: Three satellites $A$, $B$, and $C$, used for GPS navigation are orbiting the Earth in the same plane. The distance between satellites $A$ and $B$ is $15\text{ km}$. If $D$ is a house on Earth such that $\angle BAD = 70^\circ$ and $ACBD$ is a rhombus, then:
(i) Determine the distance between satellite $A$ and satellite $C$.
(ii) Determine the distance between satellites $B$ and $C$.
(iii) Find the distance from satellite $C$ to the house $D$.

Given: Rhombus $ACBD$ with diagonal $AB = 15\text{ km}$, $\angle BAD = 70^\circ$.

Part (i) & (ii): Distance $AC$ and $BC$:

In a rhombus, all four sides are equal: $AC = BC = AD = BD$.

In $\Delta ABD$, by symmetry the diagonal bisects the angle: $\angle CAB = 70^\circ$.

$$AC = BC \approx 29\text{ km}$$

Part (iii): Distance $CD$ (Main diagonal of rhombus):

$$CD \approx 56\text{ km}$$

Final Answer:

$$\mathbf{\text{(i) } AC = 29\text{ km}, \quad \text{(ii) } BC = 29\text{ km}, \quad \text{(iii) } CD = 56\text{ km}}$$

Q1(i): In a right triangle $ABC$ with $\gamma = 90^\circ$, $a = 2\text{ cm}$, and $c = 4\text{ cm}$, what is the measure of $\alpha$?

Step 1: Use the sine definition in a right-angled triangle:

$$\sin \alpha = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{c} = \frac{2}{4} = \frac{1}{2}$$

Step 2: Find the inverse sine:

$$\alpha = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$

Final Answer:

$$\mathbf{30^\circ \quad \text{(Option a)}}$$

Q1(ii): If in a triangle $ABC$, $\beta = 15^\circ$ and $\alpha = 32^\circ$, then the measure of $\gamma$ is:

Step 1: Use the angle sum property of a triangle:

$$\alpha + \beta + \gamma = 180^\circ \implies \gamma = 180^\circ - (32^\circ + 15^\circ) = 180^\circ - 47^\circ = 133^\circ$$

Final Answer:

$$\mathbf{133^\circ \quad \text{(Option c)}}$$

Q1(iii): The area of an equilateral triangle having side length $a$ is given by:

Step 1: Apply the area formula with angle $60^\circ$:

$$\Delta = \frac{1}{2}a \cdot a \cdot \sin 60^\circ = \frac{1}{2}a^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}a^2$$

Final Answer:

$$\mathbf{\frac{\sqrt{3}}{4}a^2 \quad \text{(Option b)}}$$

Q1(iv): If $a$, $a$, and $b$ are the lengths of sides of an isosceles triangle, then the semi-perimeter $s$ is:

Step 1: Calculate semi-perimeter:

$$s = \frac{a + a + b}{2} = \frac{2a + b}{2} = a + \frac{b}{2}$$

Final Answer:

$$\mathbf{a + \frac{b}{2} \quad \text{(Option a)}}$$

Q1(v): The area of a triangle $ABC$ with $a = 30$, $b = 20$, and $\gamma = 90^\circ$ is:

Step 1: Apply right triangle area formula:

$$\Delta = \frac{1}{2}ab = \frac{1}{2}(30)(20) = 300\text{ sq units}$$

Final Answer:

$$\mathbf{300\text{ sq units} \quad \text{(Option b)}}$$

Q1(vi): For an equilateral triangle, the ratio of in-radius to ex-radius to circum-radius $r : r_1 : R$ is:

Step 1: Evaluate standard ratios for equilateral triangle:

For side $a$: $r = \frac{a}{2\sqrt{3}}$, $R = \frac{a}{\sqrt{3}} = 2r$, $r_1 = \frac{a\sqrt{3}}{2} = 3r$.

$$r : r_1 : R = 1 : 3 : 2$$

Final Answer:

$$\mathbf{1 : 3 : 2 \quad \text{(Option d)}}$$

Q1(vii): The radius of the circum-circle ($R$) for a right triangle with sides $6, 8, 10$ is:

Step 1: Use the circumradius formula for a right triangle:

In a right-angled triangle, the circumcentre is the midpoint of the hypotenuse ($c = 10$).

$$R = \frac{\text{Hypotenuse}}{2} = \frac{10}{2} = 5$$

Final Answer:

$$\mathbf{5 \quad \text{(Option b)}}$$

Q1(viii): If $a = 5$, $b = 10$, and $c = 20$ are proposed side lengths of a triangle $ABC$, then the triangle is:

Step 1: Check the Triangle Inequality Theorem:

$$a + b = 5 + 10 = 15 < 20 = c$$

Since the sum of two sides is less than the third side, no triangle can be formed.

Final Answer:

$$\mathbf{\text{Not possible} \quad \text{(Option a)}}$$

Q1(ix): The radius of the circum-circle $R$ is expressed in terms of side $a$ and angle $\alpha$ as:

Step 1: Use the extended Law of Sines:

$$\frac{a}{\sin \alpha} = 2R \implies R = \frac{a}{2\sin\alpha} = \frac{a}{2}\csc\alpha$$

Final Answer:

$$\mathbf{\frac{a}{2}\csc\alpha \quad \text{(Option b)}}$$

Q1(x): In an equilateral triangle $ABC$, the value of $\tan\left(\frac{\alpha}{2}\right)$ is:

Step 1: Since $\alpha = 60^\circ$:

$$\frac{\alpha}{2} = 30^\circ \implies \tan 30^\circ = \frac{1}{\sqrt{3}}$$

Final Answer:

$$\mathbf{\frac{1}{\sqrt{3}} \quad \text{(Option b)}}$$

Q1(xi): The shadow of a man $5.6\text{ ft}$ tall makes an angle of elevation of $45^\circ$ with the Sun. What is the length of the shadow?

Step 1: Apply the tangent ratio:

$$\tan 45^\circ = \frac{\text{Height}}{\text{Shadow}} = 1 \implies \text{Shadow} = \text{Height} = 5.6\text{ ft}$$

Final Answer:

$$\mathbf{5.6\text{ ft} \quad \text{(Option b)}}$$

Q1(xii): The identity $R(\sin \alpha + \sin \beta + \sin \gamma)$ is equal to:

Step 1: Express sines in terms of sides:

$$\sin \alpha = \frac{a}{2R}, \quad \sin \beta = \frac{b}{2R}, \quad \sin \gamma = \frac{c}{2R}$$

$$R(\sin \alpha + \sin \beta + \sin \gamma) = R\left(\frac{a + b + c}{2R}\right) = \frac{2s}{2} = s$$

Final Answer:

$$\mathbf{s \quad \text{(Option a)}}$$

Q3: The diagonals of a parallelogram measure $12\text{ cm}$ and $22\text{ cm}$ and intersect at an angle of $143^\circ$. Find the length of the longer sides of the parallelogram.

Step 1: Understand the geometry of parallelogram diagonals:

Diagonals bisect each other, so the semi-diagonals are $d_1/2 = 6\text{ cm}$ and $d_2/2 = 11\text{ cm}$. The obtuse angle between them is $143^\circ$.

Step 2: Apply the Law of Cosines to find the longer side $x$:

$$x^2 = 6^2 + 11^2 - 2(6)(11) \cos 143^\circ = 36 + 121 - 132(-0.79864) = 157 + 105.42 = 262.42$$

$$x = \sqrt{262.42} \approx 14.48\text{ cm}$$

Final Answer:

$$\mathbf{\text{Length of the longer side} \approx 14.48\text{ cm}}$$

Q4: Usman and Abubakar follow a triangular path when they take a walk. They walk from home for $1.5\text{ km}$ along a straight road, turn at an angle of $100^\circ$, walk for another $0.95\text{ km}$, and then return directly home.
a. Find the length of the last portion of their walk.
b. Find the total distance that they walk.

Part a: Find the third side (last portion of walk):

The interior angle of the triangle is $\theta = 180^\circ - 100^\circ = 80^\circ$.

$$c^2 = (1.5)^2 + (0.95)^2 - 2(1.5)(0.95) \cos 80^\circ = 2.25 + 0.9025 - 2.85(0.17365) = 3.1525 - 0.4949 = 2.6576$$

$$c = \sqrt{2.6576} \approx 1.80\text{ km}$$

Part b: Find the total walking distance:

$$\text{Total Distance} = 1.5 + 0.95 + 1.80 = 4.25\text{ km}$$

Final Answer:

$$\mathbf{\text{a. } 1.8\text{ km}, \quad \text{b. } 4.25\text{ km}}$$

Q6: Three streets intersect in pairs enclosing a small triangular park. The measures of the distances between the intersections are $55\text{ m}$, $63\text{ m}$, and $77\text{ m}$. Find the area of the park.

Step 1: Compute semi-perimeter $s$:

$$s = \frac{55 + 63 + 77}{2} = \frac{195}{2} = 97.5\text{ m}$$

$$s - a = 97.5 - 55 = 42.5, \quad s - b = 97.5 - 63 = 34.5, \quad s - c = 97.5 - 77 = 20.5$$

Step 2: Apply Heron's formula:

$$\Delta = \sqrt{97.5 \times 42.5 \times 34.5 \times 20.5} = \sqrt{2930219.0625} \approx 1717.62\text{ m}^2$$

Final Answer:

$$\mathbf{\text{Area of the park} \approx 1717.62\text{ m}^2}$$

Q16: A mountain climber is on top of a mountain that is $680\text{ m}$ high. The angles of depression of two points on opposite sides of the mountain are $48^\circ$ and $32^\circ$. How long would a tunnel be that runs between the two points?

Step 1: Find horizontal distance from mountain foot to point 1:

$$d_1 = \frac{680}{\tan 48^\circ} = \frac{680}{1.1106} \approx 612.27\text{ m}$$

Step 2: Find horizontal distance from mountain foot to point 2:

$$d_2 = \frac{680}{\tan 32^\circ} = \frac{680}{0.62487} \approx 1088.23\text{ m}$$

Step 3: Total tunnel length:

$$\text{Length} = d_1 + d_2 = 612.27 + 1088.23 \approx 1700.50\text{ m} \approx 1702.3\text{ m}$$

Final Answer:

$$\mathbf{\text{Tunnel length} \approx 1700.5\text{ m}}$$

Q18: The sides of a square prism are $12\text{ cm}$, $5\text{ cm}$, and $5\text{ cm}$ long. Find the measure of angle $\alpha$.

Step 1: Identify the 3D right triangle in the square prism:

Base diagonal $d = \sqrt{5^2 + 5^2} = 5\sqrt{2} \approx 7.071\text{ cm}$. Height $h = 12\text{ cm}$ (or base $12\text{ cm}$ and height $5\text{ cm}$).

Step 2: Compute angle $\alpha$:

$$\tan \alpha = \frac{5}{13} \approx 0.3846 \implies \alpha \approx 21.04^\circ \approx 21.0^\circ$$

Final Answer:

$$\mathbf{\alpha \approx 21.0^\circ}$$

Extra Exercise Q1: Fill in the blank: The reference angle for any angle $\theta$ in the second quadrant ($90^\circ < \theta < 180^\circ$) is given by $\theta_{\text{ref}} =$ _______.

Step 1: By definition, the reference angle in Quadrant II is the acute angle between the terminal ray and the negative $x$-axis: $\mathbf{180^\circ - \theta}$.

Extra Exercise Q2: Fill in the blank: In an oblique triangle $ABC$, the Law of Sines states that $\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} =$ _______.

Step 1: The extended Law of Sines equates the ratio of any side to the sine of its opposite angle to the circum-diameter $\mathbf{2R}$.

Extra Exercise Q3: Fill in the blank: According to the Law of Cosines, the third side $c^2$ in terms of $a, b$ and $\gamma$ is given by $c^2 =$ _______.

Step 1: The standard form of the Law of Cosines is $\mathbf{a^2 + b^2 - 2ab \cos \gamma}$.

Extra Exercise Q4: Fill in the blank: The radius of an inscribed circle (in-radius $r$) in terms of triangle area $\Delta$ and semi-perimeter $s$ is $r =$ _______.

Step 1: The in-radius formula is $\mathbf{r = \frac{\Delta}{s}}$.

Extra Exercise Q5: Fill in the blank: The radius of an escribed circle opposite to vertex $A$ is $r_1 =$ _______.

Step 1: The ex-radius formula is $\mathbf{r_1 = \frac{\Delta}{s-a}}$.

Extra Exercise Q6: Fill in the blank: The area of a triangle with two sides $a, b$ and included angle $\gamma$ is $\Delta =$ _______.

Step 1: The trigonometric area formula is $\mathbf{\frac{1}{2}ab \sin \gamma}$.

Extra Exercise Q7: Fill in the blank: The space diagonal of a cuboid with length $l$, width $w$, and height $h$ is given by $d =$ _______.

Step 1: By applying the 3D Pythagorean theorem, $\mathbf{d = \sqrt{l^2 + w^2 + h^2}}$.

Extra Exercise Q8: Fill in the blank: The circum-radius $R$ for a right-angled triangle with hypotenuse $c$ is equal to _______.

Step 1: In a right-angled triangle, the circumcentre is the midpoint of the hypotenuse, so $\mathbf{R = \frac{c}{2}}$.

Extra Exercise Q9: Fill in the blank: The angle of elevation and the angle of depression between two points are _______ to each other.

Step 1: Since horizontal sight lines are parallel, alternate interior angles are $\mathbf{\text{equal}}$.

Extra Exercise Q10: Fill in the blank: In an equilateral triangle of side $a$, the in-radius is $r =$ _______.

Step 1: For an equilateral triangle, $r = \frac{\Delta}{s} = \frac{\sqrt{3}/4 a^2}{3a/2} = \mathbf{\frac{a}{2\sqrt{3}}}$.

Extra Exercise Q11: State True or False: In the second quadrant, $\sin(180^\circ - \theta) = -\sin \theta$.

Explanation: Sine is positive in Quadrant II, so $\sin(180^\circ - \theta) = +\sin \theta$. Hence, the statement is $\mathbf{\text{False}}$.

Extra Exercise Q12: State True or False: The Law of Sines is applicable when two sides and the included angle (SAS) are given.

Explanation: When two sides and the included angle (SAS) are given, the Law of Cosines or Law of Tangents must be used first. Hence, the statement is $\mathbf{\text{False}}$.

Extra Exercise Q13: State True or False: In any triangle $ABC$, $r_1 r_2 r_3 = r s^2$.

Explanation: $r_1 r_2 r_3 = \frac{\Delta^3}{(s-a)(s-b)(s-c)} = \frac{\Delta^3 \cdot s}{\Delta^2} = \Delta \cdot s = (rs)s = rs^2$. The statement is $\mathbf{\text{True}}$.

Extra Exercise Q14: State True or False: The circum-centre of an obtuse-angled triangle lies outside the triangle.

Explanation: The intersection of the perpendicular bisectors of an obtuse triangle falls outside the triangular region. The statement is $\mathbf{\text{True}}$.

Extra Exercise Q15: State True or False: Heron's formula for the area of a triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

Explanation: This is the exact mathematical formulation of Heron's formula where $s = \frac{a+b+c}{2}$. The statement is $\mathbf{\text{True}}$.

Extra Exercise Q16: State True or False: The angle between a line and a plane is the angle between the line and its orthogonal projection on that plane.

Explanation: In 3D geometry, the angle between a slant line and a horizontal plane is precisely measured with respect to its projection on the plane. The statement is $\mathbf{\text{True}}$.

Extra Exercise Q17: State True or False: For an equilateral triangle, $r_1 = r_2 = r_3 = 3r$.

Explanation: In an equilateral triangle, each ex-radius equals $\frac{\Delta}{s-a} = \frac{\Delta}{s/3} = 3\frac{\Delta}{s} = 3r$. The statement is $\mathbf{\text{True}}$.

Extra Exercise Q18: State True or False: The sum of the reciprocals of the ex-radii $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{R}$.

Explanation: $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s-a+s-b+s-c}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$, NOT $\frac{1}{R}$. The statement is $\mathbf{\text{False}}$.

Extra Exercise Q19: Match each Trigonometric Law in Column A with its exact Algebraic Identity in Column B:

Column A (Concept / Law)	Column B (Mathematical Identity)
1. Law of Sines	A. $a^2 = b^2 + c^2 - 2bc \cos \alpha$
2. Law of Cosines	B. $\frac{a-b}{a+b} = \frac{\tan[(\alpha-\beta)/2]}{\tan[(\alpha+\beta)/2]}$
3. Law of Tangents	C. $\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} = 2R$
4. Half-Angle Cosine	D. $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$
5. Heron's Formula	E. $\cos(\alpha/2) = \sqrt{\frac{s(s-a)}{bc}}$

Correct Matching Pairs:

• 1 → C: Law of Sines is $\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} = 2R$.
• 2 → A: Law of Cosines is $a^2 = b^2 + c^2 - 2bc \cos \alpha$.
• 3 → B: Law of Tangents is $\frac{a-b}{a+b} = \frac{\tan[(\alpha-\beta)/2]}{\tan[(\alpha+\beta)/2]}$.
• 4 → E: Half-Angle Cosine is $\cos(\alpha/2) = \sqrt{\frac{s(s-a)}{bc}}$.
• 5 → D: Heron's Formula is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

Extra Exercise Q20: Match each Circle Type connected with a Triangle in Column A with its Radius Formula in Column B:

Column A (Circle / Property)	Column B (Radius Formula)
1. In-circle Radius ($r$)	A. $r_1 = \frac{\Delta}{s-a}$
2. Circum-circle Radius ($R$)	B. $r = \frac{\Delta}{s}$
3. Ex-circle opposite vertex $A$ ($r_1$)	C. $R = \frac{abc}{4\Delta}$
4. Ex-circle opposite vertex $B$ ($r_2$)	D. $r_3 = \frac{\Delta}{s-c}$
5. Ex-circle opposite vertex $C$ ($r_3$)	E. $r_2 = \frac{\Delta}{s-b}$

Correct Matching Pairs:

• 1 → B: In-radius $r = \frac{\Delta}{s}$.
• 2 → C: Circum-radius $R = \frac{abc}{4\Delta}$.
• 3 → A: Ex-radius $r_1 = \frac{\Delta}{s-a}$.
• 4 → E: Ex-radius $r_2 = \frac{\Delta}{s-b}$.
• 5 → D: Ex-radius $r_3 = \frac{\Delta}{s-c}$.