Mastery Guide: Application of Trigonometry, Oblique Triangles, Connected Circles & 3D Spatial Geometry

🎯 The "All Students Take Calculus" (ASTC) Rule

Remember the signs of trigonometric ratios across quadrants I, II, III, and IV: All positive (Q1) → Sine positive (Q2) → Tangent positive (Q3) → Cosine positive (Q4).

📐 Law Selector Strategy

If you know 3 sides (SSS) or 2 sides + included angle (SAS), start with the Law of Cosines! If you know 2 angles + 1 side (AAS/ASA), use the Law of Sines!

🔍 Alternate Interior Sight Lines

The Angle of Elevation from an observer on the ground to an object equals the Angle of Depression from the object looking back down at the observer!

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

$$\mathbf{-0.559}$$

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:

$$\mathbf{0.83}$$

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:

$$\mathbf{-1.485}$$

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:

$$\mathbf{-0.674}$$

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:

$$\mathbf{-1.789}$$

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:

$$\mathbf{1.204}$$

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

Step 1: Calculate the flagpole height:

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

Final Answer:

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

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:

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

Step 1: Compute tree height:

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

Final Answer:

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

Step 1: Apply the tangent relationship:

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

Final Answer:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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

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 $a + b + c = 2s$:

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

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

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 $r \cdot s$:

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

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

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:

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

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

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

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

Step 1: Calculate semi-perimeter:

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

Final Answer:

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

Step 1: Apply right triangle area formula:

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

Final Answer:

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

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:

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

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:

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

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:

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

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:

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

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

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

Final Answer:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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 $\mathbf{\text{False}}$.

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 $\mathbf{\text{True}}$.

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

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

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 $\mathbf{\text{True}}$.

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 $\mathbf{\text{True}}$.

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 $\mathbf{\text{False}}$.

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 $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

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 $r_3 = \frac{\Delta}{s-c}$.