Mastery Guide: Trigonometry & Bearing — Angle Systems, Circle Sectors, Unit Circle Ratios, Fundamental Identities, Real-World Heights & Distances, and 3-Digit True Bearings
Mastery Guide: Trigonometry & Bearing
Single National Curriculum (SNC) • Angle Systems, Circular Measure, Trigonometric Ratios & Identities, Elevation/Depression & 3-Digit True Bearings
📖 1. Unit Overview & Target Learning Outcomes
Trigonometry (from Greek trigōnon "triangle" and metron "measure") is the branch of mathematics that connects angular rotation with linear distances. From measuring astronomical distances to guiding aircraft through crosswinds, trigonometry and directional bearings provide the universal coordinate language for modern navigation, engineering, and architecture.
🎯 Core Learning Outcomes:
- Systems of Angle Measurement: Master conversions between the Sexagesimal System (Degrees, Minutes, Seconds: $D^\circ M' S''$) and the Circular System (Radians).
- Arc Length & Sector Area: Derive and apply the fundamental circular formulas $l = r\theta$ and $A = \frac{1}{2}r^2\theta$, strictly ensuring $\theta$ is in radians.
- General Trigonometric Ratios & ASTC Rule: Evaluate the 6 trigonometric ratios ($\sin, \cos, \tan, \csc, \sec, \cot$) on the coordinate plane using standard position and the All-Silver-Tea-Cups (ASTC) quadrant sign rule.
- Special & Quadrantal Angles: Calculate exact trigonometric values for standard angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$).
- Fundamental Trigonometric Identities: Prove and manipulate Pythagorean identities ($\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$) and solve right-angled triangles.
- Heights & Distances: Model real-world scenarios using angles of elevation and depression from horizontal reference lines.
- Directional Bearings: Construct and calculate 3-digit true bearings measured clockwise from North ($000^\circ$ to $360^\circ$), determining reverse/back bearings and multi-leg navigation pathways.
💡 2. Kid-Friendly Tips for Success & Memory Hooks
☕ ASTC Rule (All Silver Tea Cups)
Remember quadrant signs: Q-I: All positive • Q-II: Sine (+ $\csc$) • Q-III: Tan (+ $\cot$) • Q-IV: Cos (+ $\sec$).
🧭 The 3-Digit Bearing Law
Bearings are always 3 digits, start at North ($000^\circ$), and measure clockwise. For example, East is $090^\circ$, not $90^\circ$; North-East is $045^\circ$!
⚠️ The Radian Alert in $l = r\theta$
In $l = r\theta$ and $A = \frac{1}{2}r^2\theta$, angle $\theta$ MUST be in radians! If given in degrees, convert first by multiplying with $\frac{\pi}{180^\circ}$.
📐 SOH - CAH - TOA & Horizons
$\sin = \frac{\text{Opp}}{\text{Hyp}}$, $\cos = \frac{\text{Adj}}{\text{Hyp}}$, $\tan = \frac{\text{Opp}}{\text{Adj}}$. Both elevation & depression angles MUST start from the horizontal line of sight!
🌍 3. Real-World Applications of Trigonometry & Bearings
✈️ Aviation & Flight Navigation
Air traffic controllers direct aircraft using exact 3-figure bearings (e.g., "Runway heading $240^\circ$"). Pilots resolve wind velocity vectors with trigonometry to stay on course.
🏗️ Civil Engineering & Surveying
Surveyors use theodolites to measure elevation angles to mountain peaks and building summits, computing immense heights without ever climbing them.
📡 Satellite Communications & Radar
Dish antennas calculate azimuth (bearing) and elevation angles to lock onto geostationary communication satellites orbiting 35,786 km above Earth.
🔑 4. Study Cues & Essential Inquiries
A radian is the ratio of two lengths: $\theta = \frac{\text{arc length}}{\text{radius}} = \frac{l}{r} \left(\frac{\text{meters}}{\text{meters}}\right)$. The units cancel out completely, making radians the natural geometric measure for physical laws.
On the unit circle, at $90^\circ$, the terminal ray lands on $(0, 1)$ where $x = 0$. Since $\tan\theta = \frac{y}{x} = \frac{1}{0}$ and $\sec\theta = \frac{1}{x} = \frac{1}{0}$, division by zero is mathematically undefined ($\infty$).
Use the $180^\circ$ parallel line rule: If bearing $\theta < 180^\circ$, $\text{Back Bearing} = \theta + 180^\circ$. If $\theta \ge 180^\circ$, $\text{Back Bearing} = \theta - 180^\circ$.
🌟 5. Section-by-Section Explanations & Visual Models
6.1 Measurement of an Angle: Sexagesimal & Circular Systems
An angle is formed by rotating a ray from an initial side to a terminal side about a common endpoint called the vertex. Anti-clockwise rotation generates positive angles; clockwise rotation generates negative angles.
• $1^\circ = 60' \text{ (minutes)} = 3600'' \text{ (seconds)}$
• Degrees to Radians: $\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180^\circ} \approx \theta_{\text{deg}} \times 0.01745\text{ rad}$
• Radians to Degrees: $\theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180^\circ}{\pi} \approx \theta_{\text{rad}} \times 57.2958^\circ$
6.2 Sector of a Circle: Arc Length & Sector Area
By direct geometric proportion, arc length $l$ and sector area $A$ are related to the total circumference $2\pi r$ and circle area $\pi r^2$ by the ratio of the central angle $\theta$ to the full revolution $2\pi$:
$$\frac{l}{2\pi r} = \frac{\theta}{2\pi} \implies \mathbf{l = r\theta}$$
$$\frac{A}{\pi r^2} = \frac{\theta}{2\pi} \implies \mathbf{A = \frac{1}{2}r^2\theta = \frac{1}{2}rl}$$
6.3 General Trigonometric Ratios & The ASTC Coordinate System
For any point $P(x, y)$ on the terminal ray at distance $r = \sqrt{x^2 + y^2} > 0$: $$\sin\theta = \frac{y}{r}, \quad \cos\theta = \frac{x}{r}, \quad \tan\theta = \frac{y}{x} \quad (x \neq 0)$$ $$\csc\theta = \frac{r}{y}, \quad \sec\theta = \frac{r}{x}, \quad \cot\theta = \frac{x}{y}$$
6.4 Fundamental Trigonometric Identities
| Identity Class | Standard Equations | Transformed Forms |
|---|---|---|
| Pythagorean Identities | $\sin^2\theta + \cos^2\theta = 1$ $1 + \tan^2\theta = \sec^2\theta$ $1 + \cot^2\theta = \csc^2\theta$ |
$\sin^2\theta = 1 - \cos^2\theta$ $\sec^2\theta - \tan^2\theta = 1$ $\csc^2\theta - \cot^2\theta = 1$ |
| Reciprocal Identities | $\csc\theta = \frac{1}{\sin\theta}, \sec\theta = \frac{1}{\cos\theta}, \cot\theta = \frac{1}{\tan\theta}$ | $\sin\theta\csc\theta = 1, \cos\theta\sec\theta = 1, \tan\theta\cot\theta = 1$ |
| Quotient Identities | $\tan\theta = \frac{\sin\theta}{\cos\theta}$ | $\cot\theta = \frac{\cos\theta}{\sin\theta}$ |
6.5 Angles of Elevation and Depression
The line of sight is the straight line drawn from the observer's eye to the target object:
• Angle of Elevation: When looking upward above the horizontal line of sight.
• Angle of Depression: When looking downward below the horizontal line of sight.
• Crucial Geometric Property: By alternate interior angles between parallel horizontals, $\text{Angle of Depression from Top} = \text{Angle of Elevation from Bottom}$.
6.6 Directional Bearings: 3-Digit True Navigation Compass
A bearing is an angle in degrees measured:
1. Strictly clockwise.
2. Starting from the North line ($000^\circ$).
3. Written as a 3-digit number (e.g., $045^\circ, 090^\circ, 180^\circ, 270^\circ, 315^\circ$).
🎯 6. Unit Synthesis Summary
Chapter 6 establishes the complete bridge between linear measurement, rotational angles, and geometric problem-solving. Angles are quantified using sexagesimal ($D^\circ M' S''$) or circular radian measures, where $1\text{ rad} = \frac{180^\circ}{\pi} \approx 57.2958^\circ$. Radians enable the direct calculation of circular arc lengths ($l = r\theta$) and sector areas ($A = \frac{1}{2}r^2\theta$). The 6 standard trigonometric ratios are generalized to coordinate planes using the ASTC quadrant rule and anchored by Pythagorean identities ($\sin^2\theta + \cos^2\theta = 1$). Finally, right-triangle trigonometry and 3-digit true bearings enable exact modeling of real-world physical heights, distances, aircraft flight paths, and maritime navigation vectors.
📝 7. Complete Solved Textbook Exercises
Exercise 6.1 • Angle Systems & Sexagesimal-Circular Conversions
Q1. Convert the following measure of angles in seconds:
Q2. Convert the following measure of angles in minutes:
Q3. Convert the following measure of angles in degrees (correct to 4 decimal places):
Q4. Write the following measures of angles in $D^\circ M' S''$:
Q5. Convert the following in radians (in terms of $\pi$):
Q6. Convert the following in radians (Use $\pi = 3.1416$):
Q7. Write the following radian measures of angles in $D^\circ M' S''$:
Exercise 6.2 • Arc Length & Sector Area
Q1. Find length of arc ($l$) and area of sector ($A$) for given radius and central angle:
Q2. A central angle in a circle of radius $4\text{ cm}$ is $75^\circ$. Find arc length and sector area.
$\theta = 75 \times \frac{\pi}{180} = \frac{5\pi}{12}\text{ rad}$.
$l = r\theta = 4 \times \frac{5\pi}{12} = \frac{5\pi}{3} \approx$ $5.24\text{ cm}$.
$A = \frac{1}{2}r^2\theta = \frac{1}{2}(16)\left(\frac{5\pi}{12}\right) = \frac{10\pi}{3} \approx$ $10.47\text{ cm}^2$.
Q3. Find radian measure of central angle ($\theta$):
Q4. Find radius ($r$) of circle:
Q5. A 30 inch pendulum swings through $30^\circ$. Find arc length.
$r = 30\text{ in}, \theta = 30^\circ = \frac{\pi}{6}\text{ rad} \implies l = r\theta = 30 \times \frac{\pi}{6} = 5\pi \approx$ $15.71\text{ inches}$.
Q6. Motorcycle on circular highway curve ($r = 10\text{ km}$, speed $42\text{ km/h}$, time $21\text{ min}$). Find angle in degrees.
Distance $l = 42 \times \frac{21}{60} = 14.7\text{ km} \implies \theta = \frac{14.7}{10} = 1.47\text{ rad} = 1.47 \times \frac{180^\circ}{\pi} \approx$ $84.22^\circ \text{ (or } 84^\circ 13' 23''\text{)}$.
Q7. Find perimeter and area of half circle with diameter $d = 12\text{ cm}$.
$r = 6\text{ cm}, \theta = \pi\text{ rad} \implies l = 6\pi \approx 18.85\text{ cm}$.
Perimeter $= l + d = 6\pi + 12 \approx$ $30.85\text{ cm}$.
Area $A = \frac{1}{2}(36)\pi = 18\pi \approx$ $56.55\text{ cm}^2$.
Q8. Find circular measure (radians) between clock hands at:
Exercise 6.3 • Trigonometric Ratios & Quadrant Rules
Q1. Complete Table of Exact Trigonometric Ratios for Standard & Quadrantal Angles:
| $\theta^\circ$ | $\theta\text{ rad}$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ | $\csc\theta$ | $\sec\theta$ | $\cot\theta$ |
|---|---|---|---|---|---|---|---|
| $0^\circ$ | $0$ | $0$ | $1$ | $0$ | Undefined | $1$ | Undefined |
| $30^\circ$ | $\frac{\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}}$ | $2$ | $\frac{2}{\sqrt{3}}$ | $\sqrt{3}$ |
| $45^\circ$ | $\frac{\pi}{4}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{\sqrt{2}}$ | $1$ | $\sqrt{2}$ | $\sqrt{2}$ | $1$ |
| $60^\circ$ | $\frac{\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | $\frac{2}{\sqrt{3}}$ | $2$ | $\frac{1}{\sqrt{3}}$ |
| $90^\circ$ | $\frac{\pi}{2}$ | $1$ | $0$ | Undefined | $1$ | Undefined | $0$ |
| $180^\circ$ | $\pi$ | $0$ | $-1$ | $0$ | Undefined | $-1$ | Undefined |
| $270^\circ$ | $\frac{3\pi}{2}$ | $-1$ | $0$ | Undefined | $-1$ | Undefined | $0$ |
| $360^\circ$ | $2\pi$ | $0$ | $1$ | $0$ | Undefined | $1$ | Undefined |
Q2. Evaluate:
Q3. For $\theta = \frac{\pi}{6} = 30^\circ$, verify identities:
- (i) $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$: LHS $= \tan 60^\circ = \sqrt{3}$; RHS $= \frac{2(1/\sqrt{3})}{1 - 1/3} = \sqrt{3}$. (Verified)
- (ii) $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$: LHS $= \sin 90^\circ = 1$; RHS $= 3(1/2) - 4(1/8) = 1$. (Verified)
- (iii) $\cos 2\theta = 1 - 2\sin^2\theta$: LHS $= \cos 60^\circ = 1/2$; RHS $= 1 - 2(1/4) = 1/2$. (Verified)
- (iv) $\sec^2\theta - \tan^2\theta = 1$: $(2/\sqrt{3})^2 - (1/\sqrt{3})^2 = 4/3 - 1/3 = 1$. (Verified)
- (v) $\csc^2\theta = 1 + \cot^2\theta$: $2^2 = 1 + (\sqrt{3})^2 \implies 4 = 4$. (Verified)
Q4. Find $\theta$ ($0 < \theta < 2\pi$):
Q5–Q7. Quadrant Sign Analysis:
Exercise 6.4 • Identities & Solving Right Triangles
Q1. Prove the following identities:
Q2–Q5. Structural Problems & Solving Triangles:
Exercise 6.5 • Applications: Heights & Distances
Exercise 6.6 • Directional Bearings & Navigation Vectors
Review Exercise 6 • Comprehensive Examination Problem Set
Q1. Select the correct option for each of the following Multiple Choice Questions:
(i) $\pi\text{ radians} = $
Explanation: By definition of circular measure, a straight angle spanning half a revolution equals $\pi\text{ radians} = 180^\circ$.
(ii) $1\text{ radian} = $
Explanation: Since $\pi\text{ rad} = 180^\circ$, dividing both sides by $\pi$ gives $1\text{ rad} = \left(\frac{180}{\pi}\right)^\circ \approx 57.296^\circ$.
(iii) $1^\circ = $
Explanation: Dividing $180^\circ = \pi\text{ rad}$ by $180$ gives $1^\circ = \frac{\pi}{180}\text{ rad} \approx 0.01745\text{ rad}$.
(iv) The value of $\cos 0^\circ$ is:
Explanation: On the unit circle at $0^\circ$, the coordinates are $(x, y) = (1, 0)$, so $\cos 0^\circ = x/r = 1/1 = 1$.
(v) The value of $\sin 90^\circ$ is:
Explanation: At $90^\circ$, the point is $(0, 1)$ on the positive y-axis, giving $\sin 90^\circ = y/r = 1/1 = 1$.
(vi) $\tan 45^\circ = $
Explanation: In an isosceles right triangle ($45^\circ-45^\circ-90^\circ$), opposite $=$ adjacent, so $\tan 45^\circ = 1$.
(vii) $\sin^2\theta + \cos^2\theta = $
Explanation: The fundamental Pythagorean trigonometric identity states $\sin^2\theta + \cos^2\theta = 1$ for all $\theta \in \mathbb{R}$.
(viii) $1 + \tan^2\theta = $
Explanation: Dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ yields $\tan^2\theta + 1 = \sec^2\theta$.
(ix) $1 + \cot^2\theta = $
Explanation: Dividing $\sin^2\theta + \cos^2\theta = 1$ by $\sin^2\theta$ yields $1 + \cot^2\theta = \csc^2\theta$.
(x) If $\sin\theta > 0$ and $\cos\theta < 0$, then the terminal arm of $\theta$ lies in:
Explanation: Sine ($y > 0$) is positive and cosine ($x < 0$) is negative exclusively in Quadrant II.
(xi) If $\tan\theta > 0$ and $\sin\theta < 0$, then the terminal arm of $\theta$ lies in:
Explanation: Tangent is positive in Q-I and Q-III. Since sine is negative ($y < 0$), $\theta$ must lie in Quadrant III ($x < 0, y < 0$).
(xii) The length of arc of a circle of radius $r$ subtending central angle $\theta$ (in radians) is:
Explanation: By definition of circular radian measure $\theta = \frac{l}{r} \implies l = r\theta$.
(xiii) The area of a circular sector of radius $r$ and central angle $\theta$ (in radians) is:
Explanation: The area of a sector is proportional to the fraction of circle: $\frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2}r^2\theta$.
(xiv) The 3-digit bearing of North is:
Explanation: In true 3-digit navigation bearing, North is the starting reference axis defined as $000^\circ$.
(xv) If the bearing of B from A is $060^\circ$, then the bearing of A from B (back bearing) is:
Explanation: For $\theta < 180^\circ$, back bearing $= \theta + 180^\circ = 060^\circ + 180^\circ = 240^\circ$.
Review Theoretical & Applied Word Problems (Q2 to Q8):
Q2. If $\cos\alpha = -\frac{5}{13}$ and the terminal arm of the angle lies in Quadrant II, find the values of $\sin\alpha, \tan\alpha,$ and $\sec\alpha$.
Step 1: In Q-II, $x = -5, r = 13 \implies y = \sqrt{r^2 - x^2} = \sqrt{169 - 25} = \sqrt{144} = +12$.
Step 2: $\sin\alpha = \frac{y}{r} = \mathbf{\frac{12}{13}}$, $\tan\alpha = \frac{y}{x} = \mathbf{-\frac{12}{5}}$, $\sec\alpha = \frac{r}{x} = \mathbf{-\frac{13}{5}}$.
Q3. A hot air balloon is rising vertically. From a point on the ground $500\text{ m}$ away, the angle of elevation changes from $30^\circ$ to $60^\circ$ in $100\text{ seconds}$. Find the upward speed of the balloon.
Step 1: Initial height $h_1 = 500\tan 30^\circ = \frac{500}{\sqrt{3}}\text{ m}$; final height $h_2 = 500\tan 60^\circ = 500\sqrt{3}\text{ m}$.
Step 2: Vertical distance $\Delta h = 500\sqrt{3} - \frac{500}{\sqrt{3}} = \frac{1000}{\sqrt{3}} \approx 577.35\text{ m}$.
Step 3: Speed $v = \frac{\Delta h}{t} = \frac{577.35}{100} \approx \mathbf{5.77\text{ m/s}}$.
Q4. In the given geometric figure, right triangle $\triangle ABD$ has height $AB = 10$, hypotenuse $AD = 16$, and inner line $AC = 14$. Find distance $a = CD$.
Step 1: In $\triangle ABD$, $BD = \sqrt{16^2 - 10^2} = \sqrt{156} \approx 12.490$.
Step 2: In $\triangle ABC$, $BC = \sqrt{14^2 - 10^2} = \sqrt{96} \approx 9.798$.
Step 3: $a = CD = BD - BC = 12.490 - 9.798 \approx \mathbf{2.69}$.
Q5. The area of a right-angled triangle is $50\text{ cm}^2$ and one of its acute angles is $45^\circ$. Find the lengths of all its sides.
Step 1: Since one acute angle is $45^\circ$, the triangle is isosceles: legs $a = b$.
Step 2: $\text{Area} = \frac{1}{2}a^2 = 50 \implies a^2 = 100 \implies a = b = \mathbf{10\text{ cm}}$.
Step 3: Hypotenuse $c = \sqrt{10^2 + 10^2} = 10\sqrt{2} \approx \mathbf{14.14\text{ cm}}$.
Q6. The shadow of a vertical building increases by $12\text{ m}$ when the angle of elevation of the sun decreases from $60^\circ$ to $45^\circ$. Find the height of the building.
Step 1: Let height be $h$. At $60^\circ$, shadow $x = \frac{h}{\sqrt{3}}$. At $45^\circ$, shadow $x + 12 = h$.
Step 2: $h - \frac{h}{\sqrt{3}} = 12 \implies h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = 12 \implies h = \frac{12\sqrt{3}}{\sqrt{3}-1} = 6(3+\sqrt{3}) \approx \mathbf{28.39\text{ m}}$.
Q7. From the top of a $100\text{ m}$ high building, an observer notices the angle of depression of the bottom of another building is $30^\circ$ and the angle of elevation of the top of the second building is $20^\circ$. Find the height of the second building.
Step 1: Horizontal distance $d = \frac{100}{\tan 30^\circ} = 100\sqrt{3} \approx 173.205\text{ m}$.
Step 2: Height of second building above observer line $h_2 = d\tan 20^\circ = 173.205(0.36397) \approx 63.04\text{ m}$.
Step 3: Total height $= 100 + 63.04 = \mathbf{163.04\text{ m}}$.
Q8. An airplane is located $119\text{ km}$ West and $100\text{ km}$ South of an airport. On what true 3-digit bearing should the pilot fly to reach the airport directly?
Step 1: Plane is at $(-119, -100)$ relative to airport. Vector to airport is $+119\text{ km}$ East, $+100\text{ km}$ North.
Step 2: Angle East of North $\theta = \tan^{-1}\left(\frac{119}{100}\right) = \tan^{-1}(1.19) \approx 49.96^\circ \approx 50^\circ$.
Step 3: 3-digit true bearing from North $= \mathbf{050^\circ}$.
More Chapter Notes for Class 9 (FBISE)
MathematicsTest Your Knowledge on Chapter 6: Mastery Guide: Trigonometry & Bearing — Angle Systems, Circle Sectors, Unit Circle Ratios, Fundamental Identities, Real-World Heights & Distances, and 3-Digit True Bearings
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