Chapter 9 · Some Applications of Trigonometry

9.1 Line of Sight, Angles of Elevation and Depression

To measure a tower you cannot climb, you only need to look at its top and measure an angle. Learn the language of the line of sight, the angle of elevation and the angle of depression, and how each one turns a real scene into a right-angled triangle.

Stand at the bottom of a tall tree and look at its topmost leaf. Your eyes have to tilt upwards. Now stand on the roof of a school building and look at a cricket ball lying on the playground: your eyes tilt downwards. In both cases you are looking along a straight line from your eye to the object, and the angle your head turns away from the level (horizontal) direction is something a student can measure with a protractor and a plastic straw — or with the clinometer app on a phone.

In this section we give names to these lines and angles, learn how to draw a neat right-angled triangle out of any scene (a tower, a kite string, a ladder, a lighthouse), and use the trigonometric ratios from Chapter 8 (trigonometric ratios) to find a height or a distance. We stay with the three standard angles $30^\circ$, $45^\circ$ and $60^\circ$, and use $\sqrt3\approx1.732$ and $\sqrt2\approx1.414$ for numerical answers.

Line of sight and the two angles

📘 Definition 9.1 — Line of sight
The line of sight is the straight line drawn from the eye of the observer to the object being viewed.

To compare directions we always need a reference. In this chapter the reference is the horizontal line through the observer's eye: the level line, parallel to the ground, that you would see if you looked straight ahead at a distant horizon. A vertical object such as a tower or a pole stands at a right angle ($90^\circ$) to the ground. That right angle is what makes a triangle in every problem of this chapter.

📘 Definition 9.2 — Angle of elevation
When the object is above the horizontal line through the eye, the angle of elevation is the angle between the line of sight and the horizontal line. The observer has to raise the eyes to see the object.
📘 Definition 9.3 — Angle of depression
When the object is below the horizontal line through the eye, the angle of depression is the angle between the line of sight and the horizontal line. The observer has to lower the eyes to see the object.

Both angles are always measured from the horizontal, never from the vertical and never from the ground at some other place. Both are acute angles ($0^\circ\lt\theta\lt90^\circ$) in the problems of this chapter. If you are looking at the top of a tower, you measure the angle of elevation at your eye; if you are standing on the tower looking at the foot of a pole, you measure the angle of depression at your eye.

📐 Theorem 9.1 — Elevation and depression are equal
Let $A$ and $B$ be two points at different heights, with $A$ higher than $B$. The angle of depression of $B$ as seen from $A$ is equal to the angle of elevation of $A$ as seen from $B$.

Proof. Draw a horizontal line through $A$ and a horizontal line through $B$. These two horizontal lines are parallel. The line of sight $AB$ is a transversal crossing both. The angle of depression at $A$ (between the horizontal through $A$ and $AB$, below it) and the angle of elevation at $B$ (between the horizontal through $B$ and $BA$, above it) lie on opposite sides of the transversal and between the parallel lines: they are alternate angles, and alternate angles on parallel lines are equal. $\blacksquare$

This is a very useful fact: when a problem gives you the angle of depression from the top of a building to a car, you may put the same angle inside the triangle at the car, where it sits next to the ground. Interactive 2 below shows this "Z" shape.

⚠️ Three mistakes students make alone

1. Measuring from the vertical. The angle of elevation is not the angle between the line of sight and the tower. It is the angle between the line of sight and the horizontal. A $30^\circ$ elevation means a gentle upward look, not a steep one.

2. Putting the depression angle at the wrong corner. The angle of depression is measured below the horizontal at the observer. The triangle's angle at the object (on the ground) is equal to it by Theorem 9.1; but the angle at the top of the tower in the triangle is $90^\circ$ minus the depression angle, not the depression angle itself.

3. Forgetting the eye height. The horizontal line is at the observer's eye, not at the ground. If a $1.5$ m person looks at a tower, the height of the tower is the height found from the triangle plus $1.5$ m. When a problem says "ignore the height of the observer", skip this.

From a scene to a right-angled triangle

Every problem in this chapter has three kinds of quantity: a vertical height $h$ (tower, pole, cliff, kite above the ground), a horizontal distance $x$ along the ground, and a slant length $l$ along the line of sight (a string, a ladder, a ray). They form a right-angled triangle with the angle of elevation or depression $\theta$ as one of the acute angles.

📐 Theorem 9.2 — The right-triangle toolkit for heights and distances
In a right-angled triangle with an acute angle $\theta$, where $h$ is the side opposite $\theta$ (vertical), $x$ is the side adjacent to $\theta$ (horizontal) and $l$ is the hypotenuse (line of sight), $$\sin\theta=\frac{h}{l},\qquad\cos\theta=\frac{x}{l},\qquad\tan\theta=\frac{h}{x}.$$ For the standard angles (recalled from Chapter 8: trigonometric ratios):
$\theta$$30^\circ$$45^\circ$$60^\circ$
$\sin\theta$$\tfrac12$$\tfrac1{\sqrt2}$$\tfrac{\sqrt3}2$
$\cos\theta$$\tfrac{\sqrt3}2$$\tfrac1{\sqrt2}$$\tfrac12$
$\tan\theta$$\tfrac1{\sqrt3}$$1$$\sqrt3$
🧭 Problem-solving strategy — from words to a triangle
  1. Draw a neat sketch: the ground as a horizontal line, the tower or pole as a vertical line, the observer as a dot, the line of sight joining them.
  2. Mark the given angle at the observer (or at the object, if you used Theorem 9.1), the right angle, and every known length. Write $h$ or $x$ for the unknown.
  3. Choose the ratio that links one known and one unknown side to the angle. Two vertical and horizontal sides? Use $\tan$. A slant length involved? Use $\sin$ (with the vertical side) or $\cos$ (with the horizontal side).
  4. Solve, using the exact value of the ratio first ($\sqrt3$, $\tfrac1{\sqrt3}$ …), and only then put $\sqrt3\approx1.732$.
  5. Check: does the answer look sensible? A $45^\circ$ look means height equals distance; a $60^\circ$ look means the height is about $1.7$ times the distance; a $30^\circ$ look means it is about $0.58$ times.
✏️ Example 1 — Height of a tower
An observer stands $20$ m from the foot of a tower on level ground. The angle of elevation of the top of the tower is $60^\circ$. Find the height of the tower. (Ignore the observer's height; use $\sqrt3\approx1.732$.)
Figure Let the tower be $AB$ (foot $B$, top $A$) and the observer be at $C$ with $BC=20$ m. In the right triangle $ABC$ (right angle at $B$), $\angle ACB=60^\circ$ and $AB=h$.
Ratio $h$ is opposite the angle and $20$ is adjacent to it, so use $\tan$: $\tan60^\circ=\dfrac{h}{20}$.
Solve $\sqrt3=\dfrac{h}{20}$, so $h=20\sqrt3\approx20\times1.732=34.64$ m.
Check At $60^\circ$ the height is about $1.7$ times the distance: $1.7\times20=34$ ✓.
✏️ Example 2 — A kite on a string
A kite is flying with its string $60$ m long, stretched straight and making an angle of $30^\circ$ with the level ground. Find the height of the kite above the ground. (Ignore the sag of the string and the height of the hand.)
Figure The string is the hypotenuse $l=60$ m, the height $h$ is opposite the $30^\circ$ angle at the boy.
Ratio Hypotenuse and opposite side: use $\sin$. $\sin30^\circ=\dfrac{h}{60}$.
Solve $\dfrac12=\dfrac{h}{60}$, so $h=30$ m.
Check In a $30^\circ$ triangle the side opposite the angle is half the hypotenuse ✓.
✏️ Example 3 — A boat seen from a lighthouse
From the top of a lighthouse $50$ m high, the angle of depression of a boat is $45^\circ$. Find (a) the horizontal distance of the boat from the foot of the lighthouse and (b) the length of the line of sight. (Use $\sqrt2\approx1.414$.)
Figure Let the top be $A$, the foot $B$ and the boat $C$. The angle of depression of $C$ from $A$ is $45^\circ$, so by Theorem 9.1 the angle of elevation of $A$ from $C$ is $\angle ACB=45^\circ$. Right angle at $B$, $AB=50$ m.
(a) Distance $\tan45^\circ=\dfrac{AB}{BC}$, so $1=\dfrac{50}{BC}$ and $BC=50$ m.
(b) Line of sight $\sin45^\circ=\dfrac{AB}{AC}$: $\dfrac1{\sqrt2}=\dfrac{50}{AC}$, so $AC=50\sqrt2\approx50\times1.414=70.7$ m.
Check $50^2+50^2=5000=70.7^2$ (approximately) ✓ by Pythagoras.
Interactive 1 — Walk Towards the Tower: Watch the Angle of Elevation

A tower $30$ m high stands on level ground; the observer's eye height is ignored. Drag the observer along the ground (or use the slider, or press Play to make the observer walk). The green arc is the angle of elevation $\theta$. The two buttons put the observer exactly where $\theta$ is a standard angle. Check the last readout: $\tan\theta\times x$ always gives the height $30$ m.

40.00
distance $x$ (m)—
angle of elevation $\theta$—
line of sight $l$ (m)—
$\tan\theta$ and $x\tan\theta$ (m)—
✏️ Example 4 — Finding the sun's angle
A vertical pole $6$ m high casts a shadow $2\sqrt3$ m long on level ground. Find the angle of elevation of the sun, and the length of the ray from the top of the pole to the tip of the shadow.
Figure The pole is the vertical side ($6$ m), the shadow is the horizontal side ($2\sqrt3$ m). The sun's angle $\theta$ is at the tip of the shadow.
Ratio $\tan\theta=\dfrac{6}{2\sqrt3}=\dfrac{3}{\sqrt3}=\sqrt3$. Since $\tan60^\circ=\sqrt3$, $\theta=60^\circ$.
Ray $\sin60^\circ=\dfrac{6}{l}$: $\dfrac{\sqrt3}{2}=\dfrac6l$, so $l=\dfrac{12}{\sqrt3}=4\sqrt3\approx6.93$ m.
Check $6^2+(2\sqrt3)^2=36+12=48=(4\sqrt3)^2$ ✓.
✏️ Example 5 — A ladder against a wall
A $10$ m ladder leans against a vertical wall and its top reaches a window $5\sqrt3$ m above the ground. Find the angle the ladder makes with the ground and how far its foot is from the wall.
Figure The ladder is the hypotenuse ($10$ m), the window height is the side opposite the angle $\theta$ at the foot.
Angle $\sin\theta=\dfrac{5\sqrt3}{10}=\dfrac{\sqrt3}{2}$, so $\theta=60^\circ$.
Distance $\cos60^\circ=\dfrac{x}{10}$: $\tfrac12=\dfrac x{10}$, so $x=5$ m.
Check $5^2+(5\sqrt3)^2=25+75=100=10^2$ ✓.
✏️ Example 6 — Include the observer's eye height
A student whose eyes are $1.5$ m above the ground stands $30$ m from a tower. The angle of elevation of its top is $30^\circ$. Find the height of the tower. (Use $\sqrt3\approx1.732$.)
Figure The horizontal line of sight starts at the eye, $1.5$ m above the ground. The triangle's vertical side is the part of the tower above eye level, call it $h$.
Triangle $\tan30^\circ=\dfrac{h}{30}$: $\dfrac1{\sqrt3}=\dfrac{h}{30}$, so $h=\dfrac{30}{\sqrt3}=10\sqrt3\approx17.32$ m.
Add the eye height Tower $=17.32+1.5=18.82$ m.
Check At $30^\circ$ the height above the eye is about $0.58\times30=17.3$ ✓.

Looking down: angles of depression

An angle of depression problem usually puts the observer high up — on a cliff, lighthouse, balcony, bridge or the pilot's seat of an aeroplane. Theorem 9.1 lets you move the angle to the ground, where the triangle is easy to see. Use Interactive 2 to convince yourself: change the position of the boat and compare the two readouts.

Interactive 2 — The Lighthouse and the Boat: Depression Equals Elevation

The top of a lighthouse is $40$ m above the sea. The boat sails away and towards the lighthouse (slider, or press Play). The red arc is the angle of depression at the lighthouse, measured below the dashed horizontal line. The blue arc is the angle of elevation at the boat. They are the two ends of a "Z" and are always equal. Use the buttons to put the boat at a standard angle.

60.00
depression at the top—
elevation at the boat—
line of sight (m)—
$\tan\delta=\tfrac{40}{x}$—

Exercises

I
Angles of elevation. Use $\sqrt3\approx1.732$, $\sqrt2\approx1.414$ and give lengths to two decimal places unless they are exact. Ignore the observer's height unless it is given.
  • 1 An observer is $15$ m from the foot of a tower and the angle of elevation of its top is $45^\circ$. Find the height of the tower.
  • 2 The angle of elevation of the top of a pole from a point $12$ m from its foot is $60^\circ$. Find the height of the pole.
  • 3 From a point $30$ m from the foot of a tower the angle of elevation of the top is $30^\circ$. Find the height.
  • 4 A pole $8$ m high casts a shadow on level ground when the sun's elevation is $30^\circ$. Find the length of the shadow.
  • 5 A tree casts a shadow exactly as long as its height. What is the sun's angle of elevation?
  • 6 A $14$ m ladder leans against a wall and makes an angle of $60^\circ$ with the ground. How high up the wall does it reach, and how far is its foot from the wall?
  • 7 A kite string $100$ m long is stretched straight at $30^\circ$ to the level ground. How high is the kite?
  • 8 When the sun's elevation is $30^\circ$ a tree casts a shadow $18$ m long. Find the height of the tree.
1 $\tan45^\circ=\dfrac h{15}$, $h=15$ m.
2 $\tan60^\circ=\dfrac h{12}$, $h=12\sqrt3\approx20.78$ m.
3 $\tan30^\circ=\dfrac h{30}$, $h=\dfrac{30}{\sqrt3}=10\sqrt3\approx17.32$ m.
4 $\tan30^\circ=\dfrac8s$, $s=8\sqrt3\approx13.86$ m.
5 $\tan\theta=\dfrac hh=1$, so $\theta=45^\circ$.
6 Height $=14\sin60^\circ=7\sqrt3\approx12.12$ m. Distance $=14\cos60^\circ=7$ m.
7 $h=100\sin30^\circ=50$ m.
8 $\tan30^\circ=\dfrac h{18}$, $h=\dfrac{18}{\sqrt3}=6\sqrt3\approx10.39$ m.
II
Angles of depression and mixed problems.
  • 1 From the top of a cliff $80$ m high, the angle of depression of a boat is $30^\circ$. How far is the boat from the foot of the cliff?
  • 2 In Part 1, what is the length of the line of sight from the top of the cliff to the boat?
  • 3 The angle of depression of a boat from the top of a $60$ m lighthouse is $60^\circ$. Find its horizontal distance from the lighthouse.
  • 4 From the top of a $45$ m tower the angle of depression of a car is $45^\circ$. How far is the car from the foot of the tower?
  • 5 A girl on a balcony $40$ m above the road sees a car at an angle of depression of $60^\circ$. How far is the car from the building?
  • 6 From a point on the ground, the angle of elevation of an aeroplane flying at a height of $3000$ m is $30^\circ$. Find the straight-line distance from the observer to the aeroplane.
  • 7 A girl whose hand is $1.2$ m above the ground flies a kite with $30$ m of string stretched straight at $60^\circ$ to the horizontal. Find the height of the kite above the ground.
1 By Theorem 9.1 the angle at the boat is $30^\circ$. $\tan30^\circ=\dfrac{80}{x}$, $x=80\sqrt3\approx138.56$ m.
2 $\sin30^\circ=\dfrac{80}{l}$, so $l=160$ m.
3 $\tan60^\circ=\dfrac{60}{x}$, $x=\dfrac{60}{\sqrt3}=20\sqrt3\approx34.64$ m.
4 $\tan45^\circ=\dfrac{45}{x}$, so $x=45$ m.
5 $\tan60^\circ=\dfrac{40}{x}$, $x=\dfrac{40}{\sqrt3}\approx23.09$ m.
6 $\sin30^\circ=\dfrac{3000}{l}$, so $l=6000$ m $=6$ km.
7 Height above the hand $=30\sin60^\circ=15\sqrt3\approx25.98$ m. Add $1.2$ m: $27.18$ m.
III
Finding the angle, and the eye height.
  • 1 A tower is $20$ m high and the observer stands $20\sqrt3$ m from its foot. Find the angle of elevation of the top.
  • 2 A $10$ m pole casts a shadow $\dfrac{10}{\sqrt3}$ m long. Find the sun's angle of elevation.
  • 3 A ladder $8$ m long has its foot $4$ m from a wall. Find the angle it makes with the ground and the height it reaches.
  • 4 A rope runs from the top of a $12$ m pole to a peg in the ground $12$ m from the pole's foot. Find the angle of the rope with the ground and the rope's length.
  • 5 A person whose eyes are $1.6$ m above the ground stands $20$ m from a building and sees its top at an elevation of $45^\circ$. Find the height of the building.
  • 6 A student with eye height $1.5$ m stands $30$ m from a tower and sees the top at $60^\circ$ elevation. Find the height of the tower.
1 $\tan\theta=\dfrac{20}{20\sqrt3}=\dfrac1{\sqrt3}$, so $\theta=30^\circ$.
2 $\tan\theta=\dfrac{10}{10/\sqrt3}=\sqrt3$, so $\theta=60^\circ$.
3 $\cos\theta=\dfrac48=\dfrac12$, so $\theta=60^\circ$; height $=8\sin60^\circ=4\sqrt3\approx6.93$ m.
4 $\tan\theta=\dfrac{12}{12}=1$, so $\theta=45^\circ$; rope $=12\sqrt2\approx16.97$ m.
5 Above the eye: $20\tan45^\circ=20$ m. Building $=20+1.6=21.6$ m.
6 Above the eye: $30\tan60^\circ=30\sqrt3\approx51.96$ m. Tower $=51.96+1.5=53.46$ m.
IV
Challenge.
  • 1 The line of sight from an observer to the top of a tower is $40$ m long and makes $60^\circ$ with the ground. Find the height of the tower and the distance of the observer from its foot.
  • 2 A $10$ m ladder rests against a wall at $60^\circ$ to the ground. Its foot is pulled away from the wall until the ladder makes $45^\circ$ with the ground. How far did the foot move? (Use $\sqrt2\approx1.414$.)
  • 3 Explain, with a sketch, why the angle of depression of a boat from the top of a cliff is not the angle at the top of the cliff inside the right triangle. What is that angle in the triangle, if the depression is $30^\circ$?
  • 4 The shadow of a vertical tower on level ground is $x$ m long when the sun's elevation is $45^\circ$. Find, in terms of $x$, the tower's height; and the shadow length when the elevation becomes $60^\circ$.
1 Height $=40\sin60^\circ=20\sqrt3\approx34.64$ m; distance $=40\cos60^\circ=20$ m.
2 Initially $10\cos60^\circ=5$ m from the wall; finally $10\cos45^\circ=5\sqrt2\approx7.07$ m. The foot moved $7.07-5=2.07$ m.
3 The depression is measured below the horizontal at the top, so it lies outside the triangle. Inside the triangle, the angle at the top is between the vertical side (the cliff) and the line of sight: $90^\circ-30^\circ=60^\circ$. The depression equals the angle at the boat (Theorem 9.1), which is $30^\circ$.
4 $\tan45^\circ=\dfrac h x$, so $h=x$. At $60^\circ$: $\tan60^\circ=\dfrac xs$, $s=\dfrac x{\sqrt3}\approx0.577x$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: heights-and-distances problems are drawn from a figure you must sketch yourself. Short items ask for one ratio ($1$–$2$ marks); 3-mark items add a second quantity (such as the line of sight). Case studies give a real scene and ask three connected questions. Marks go for the figure, the ratio written down, the substitution and the answer with units.

E1. Board · 1 mark A tower $100$ m high casts a shadow on level ground when the sun's elevation is $30^\circ$. The length of the shadow is  (A) $\dfrac{100}{\sqrt3}$ m  (B) $100\sqrt3$ m  (C) $50$ m  (D) $200$ m

Ratio $\tan30^\circ=\dfrac{100}{s}$, so $s=\dfrac{100}{\tan30^\circ}=100\sqrt3\approx173.2$ m.
Trap (A) $\dfrac{100}{\sqrt3}\approx57.7$ m comes from writing $s=100\tan30^\circ$, i.e. putting the shadow on top of the fraction. At a low sun ($30^\circ$) the shadow must be longer than the tower, so (A) and (C) are too short.
Answer (B).

E2. Board · 1 mark Assertion (A): When the sun's angle of elevation is $45^\circ$, a vertical pole and its shadow have the same length. Reason (R): $\tan45^\circ=1$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R $\tan45^\circ=1$ is a standard value. True.
Check A $\tan45^\circ=\dfrac{\text{pole}}{\text{shadow}}=1$, so pole $=$ shadow. True, and this is exactly the reason R.
Answer (A).

E3. Board · 2 marks A kite is at a height of $60$ m above the ground. The string attached to it is stretched straight and makes an angle of $60^\circ$ with the level ground. Find the length of the string. (Use $\sqrt3\approx1.732$.)

Ratio (1 mark) The string is the hypotenuse $l$; the height $60$ m is opposite the $60^\circ$ angle: $\sin60^\circ=\dfrac{60}{l}$.
Solve (1 mark) $\dfrac{\sqrt3}{2}=\dfrac{60}{l}$, $l=\dfrac{120}{\sqrt3}=40\sqrt3\approx40\times1.732=69.28$ m.

E4. Board · 3 marks From the top of a vertical cliff $75$ m high, the angle of depression of a car on the level road below is $30^\circ$. Find the horizontal distance of the car from the foot of the cliff and the length of the line of sight. (Use $\sqrt3\approx1.732$.)

Figure (1 mark) Cliff $AB=75$ m, car at $C$. The angle of depression is $30^\circ$, so $\angle ACB=30^\circ$ (alternate angles, Theorem 9.1).
Distance (1 mark) $\tan30^\circ=\dfrac{75}{BC}$, so $BC=75\sqrt3\approx75\times1.732=129.9$ m.
Line of sight (1 mark) $\sin30^\circ=\dfrac{75}{AC}$, so $AC=150$ m.
Check $75^2+129.9^2\approx5625+16874=22499\approx150^2$ ✓.

E5. Board · Case study A group of students in a science club measure the height of a mobile-phone tower. Each student's eye is $1.6$ m above the ground. Meera stands $18$ m from the foot of the tower and measures the angle of elevation of its top as $60^\circ$. (Use $\sqrt3\approx1.732$.)
(i) Which trigonometric ratio would you use to find the height of the top above Meera's eye, and what is that height? (2 marks)
(ii) What is the total height of the tower? (1 mark)
(iii) Kabir wants the angle of elevation to be $45^\circ$. How far from the foot of the tower should he stand? (1 mark)

(i) The opposite side (height above eye level) and the adjacent side ($18$ m) are involved, so use $\tan$: $\tan60^\circ=\dfrac{h}{18}$, $h=18\sqrt3\approx31.18$ m. (1 + 1 marks)
(ii) Tower $=31.18+1.6=32.78$ m. (1 mark)
(iii) At $45^\circ$ the horizontal distance equals the height above eye level: $31.18$ m (that is, $18\sqrt3$ m). (1 mark)
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. How accurately can you measure the height of your school building, a tree or a lamp-post using only a phone clinometer app and a tape measure?

Why it matters. Surveyors, foresters and engineers measure heights they cannot climb. Forest departments estimate timber from tree heights; telecom engineers check tower heights; the same idea, with a theodolite instead of a phone, is how the heights of mountains were first measured in India.

What to do.

  1. Choose a tall object with clear level ground in front of it. Measure your eye height $e$ with a tape.
  2. Stand at three different distances $x$ (for example $10$ m, $15$ m, $20$ m) from the foot of the object, measuring each with the tape. At each place use the clinometer app to read the angle of elevation $\theta$ of the top.
  3. For each place compute the height $h=x\tan\theta+e$. A calculator's tan key is allowed here, since the angles you measure will not be $30^\circ$, $45^\circ$ or $60^\circ$.
  4. Move until the app reads $45^\circ$ and measure $x$ again: the height above your eye should equal $x$.
  5. Draw a neat scale figure of your best measurement and write down the right-angled triangle.

Tools: A phone with a clinometer app, a tape measure (or a rope marked in metres), a notebook

Hand in: A table with columns $x$, $\theta$, $x\tan\theta$, $h$; your three estimates and their mean; a short note on why they differ.

Going further. Compare your estimate with a measurement by the shadow method (a $1$ m stick and its shadow). Which seems more reliable, and why?

Teacher/parent note: results will differ by a few percent because of errors in reading the angle and holding the phone level. Errors grow when the angle is very small or very large; angles near $45^\circ$ are the most reliable. The standard-angle values used in the lessons are for exact working; calculator values are fine for the project.

🧪 Self-check quiz

0/5 correct

Q1. The angle of elevation of the top of a $30$ m tower from a point $30$ m from its foot is

Q2. From a point A on a cliff the angle of depression of a boat B is $35^\circ$. The angle of elevation of A from B is

Q3. A kite string $40$ m long is stretched straight and makes $30^\circ$ with the level ground. The height of the kite above the ground is

Q4. A tower $15$ m high casts a shadow $15\sqrt3$ m long. The sun's angle of elevation is

Q5. From the top of a $30$ m cliff the angle of depression of a boat is $60^\circ$. The boat's horizontal distance from the foot of the cliff is

← Chapter 9 — Some Applications of Trigonometry9.2 Heights and Distances →