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Some Applications of Trigonometry
Heights and Distances
Practice: Elevation and Depression
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Some Applications of TrigonometryHeights and Distances

Reading time: ~60 min

Riya's class is on a school trip to an old stone minar. Before lunch, their maths teacher sets a challenge: find the height of the minar. Nobody may climb it, and nobody may drop a tape measure from the top.

Riya has brought a home-made clinometer — a protractor with a drinking straw taped along its edge and a thread with a small weight hanging from its centre. She looks at the top of the minar through the straw, and the thread shows her the angle.

In the previous chapter you learnt the trigonometric ratios of an angle. In this chapter you will put them to work, and find heights and distances that no tape measure can reach.

Fig. 9.1 is live. Riya's eye is at A, 1.5 m above the ground, and she stands 30 m from the foot D of the minar. Drag the end of her line of sight and aim it exactly at the top C of the minar.

The line from Riya's eye to the top of the minar, AC, is called the line of sight. The line AB runs level from her eye to the minar: it is the horizontal. The angle BAC between them is called the angle of elevation of the top of the minar from her eye. Riya's thread reads 45°.

The line of sight is the line drawn from the eye of an observer to the point being looked at.

The angle of elevation of a point is the angle that the line of sight makes with the horizontal when the point is above the horizontal level — that is, when we raise our head to look at it.

One colour code runs through every figure in this chapter: heights are green, distances along the ground and other horizontal lines are blue, lines of sight are amber, and angles are purple.

Measuring the Unreachable

Riya's trick is far older than her clinometer. In the 1800s, the Great Trigonometrical Survey of India set out to map the whole subcontinent using measured distances and angles. Surveyors hauled heavy theodolites — precise angle-measuring telescopes — up hills and towers, and sighted distant points from one station after another.

In 1852, Radhanath Sikdar, a mathematician from Kolkata working for the Survey, worked through the angles that had been measured to a far-off Himalayan summit known only as Peak XV. His result: it was the highest mountain on Earth. Today we call it Everest.

How could Sikdar find the height of a peak that nobody had ever climbed?

A rope let down from the summit
Angles of elevation measured from far-away stations, and the distances between the stations
Timing a stone dropped from the top

That is exactly Riya's toolkit: an angle you can measure, a distance you can walk, and a trigonometric ratio to join the two. It is the whole of this chapter.

Look Up, Look Down

Riya's friend Aman has climbed up to the balcony of the temple next to the minar. From there he can see a kite flying above him, and a flower pot on the temple stairs below him.

Fig. 9.2 and Fig. 9.3 are live. Aman is looking at a point P: the kite above him, or the flower pot below him. Drag P up and down and complete both missions.

When P is below Aman's eye, the is below the horizontal level. The angle it then makes with the horizontal has a name of its own: the angle of depression.

The angle of depression of a point is the angle that the line of sight makes with the horizontal when the point is below the horizontal level — that is, when we lower our head to look at it.

The previous chapter showed a few people looking at things. Here are some more. For each sighting, is the angle an angle of elevation or an angle of depression? Drag each one into the right bucket.

a pilot looking at the runway
Riya looking at the minar top
a lifeguard watching a swimmer
a fisherman looking at a lighthouse
a girl on a bridge looking at the river bank
a farmer watching a kite
Aman looking at the flower pot
a child looking at a mango on a tree
Angle of elevation
Angle of depression

Tip

Ask one question: is the object above or below the eye of the person looking? Above means an angle of elevation, below means an angle of depression. Both are always measured from the horizontal, never from the vertical.

One Angle, Two Names

Aman's uncle keeps a lighthouse on the Konkan coast. From the top he looks down at a fishing boat, and at the same moment the fisherman looks up at him. Two people, one line of sight — and two angles.

Drag the boat and watch the angle of depression at the top and the angle of elevation from the boat.

Wherever the boat goes, the angle of of the boat from the top is equal to the angle of of the top from the boat.

Here is why. The horizontal at the top and the sea level are lines, and the line of sight cuts across both of them: it is a transversal. So the two angles are angles, and alternate angles are always equal.

Tip

In almost every angle-of-depression problem, the first move is to copy the angle down to the ground, where it becomes an angle of elevation inside a right triangle. Alternate angles make that move legal.

How Tall Is the Minar?

Back to Riya. To find the height CD of the minar without climbing it, she needs three things:

(i) the distance DE from her feet to the foot of the minar,

(ii) the angle of , ∠BAC, of the top of the minar, and

(iii) her own eye height, AE.

The height of the minar splits into two parts: CD = CB + BD, and BD is just Riya's eye height, AE = 1.5 m.

To find CB, we use ratios of ∠BAC, the angle she measured.

In the right triangle ABC, the side CB is the side for the known angle A, and AB is the adjacent side. The ratios that join exactly these two sides are tan A and cot A:

\tan A = \dfrac{BC}{AB}   or   \cot A = \dfrac{AB}{BC}

Fig. 9.1 — Riya's sighting of the minar. (Riya is drawn taller than scale.)

With Riya's numbers, AB = DE = 30 m and ∠A = 45°:

\tan 45^\circ = \dfrac{BC}{30}

1 = \dfrac{BC}{30}

So BC = m, and the minar is CD = BC + 1.5 = m tall.

The method for every problem in this chapter

  1. Draw the figure and pick out the right triangle.
  2. Mark the side you know, the side you want, and the angle you know.
  3. Choose the ratio that joins those two sides: opposite and adjacent, tan; opposite and hypotenuse, sin; adjacent and hypotenuse, cos.
  4. Solve for the side you want. If the angle was measured from someone's eye, add the eye height at the end.

Tip

Learn these three by heart — nearly every problem in this chapter uses them: \tan 30^\circ = \dfrac{1}{\sqrt{3}}, \tan 45^\circ = 1 and \tan 60^\circ = \sqrt{3}. When a question says "take \sqrt{3} = 1.732", use that value only in the very last line.

Speed Round: Which Ratio?

Step 3 of the method, at speed. Each triangle shows a side you know, the side you want (marked x) and an angle. Pick the ratio that joins the two sides.

Get five right to move on.

The Tower and Its Angle

Example 1 : A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Solution : First draw the figure, Fig. 9.4. The tower AB is the height we want, the distance CB is 15 m, and the angle of elevation ∠ACB is 60°. Triangle ACB is right-angled at B.

AB is opposite the 60° angle and CB is adjacent to it, so the ratio to use is 60° (or cot 60°).

\tan 60^\circ = \dfrac{AB}{BC}

\sqrt{3} = \dfrac{AB}{15}

AB = 15\sqrt{3}

So the height of the tower is m — about 26 m.

Fig. 9.4 — the tower AB and the point C, 15 m from its foot.

The Electrician's Ladder

Example 2 : An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take \sqrt{3} = 1.73)

Fig. 9.5 is live. The pole AD is 5 m tall, and B, the point she must reach, is 1.3 m below its top. The ladder always leans at 60° to the ground. Slide the length of the ladder until its top reaches B.

Length of the ladder: ${len} m

Solution : The electrician has to reach B, so the height she needs is BD = AD − AB = (5 − 1.3) m = m.

The ladder BC is the hypotenuse of the right triangle BDC, and BD is opposite the 60° angle. So the ratio to use is 60°.

\dfrac{BD}{BC} = \sin 60^\circ

\dfrac{3.7}{BC} = \dfrac{\sqrt{3}}{2}

BC = \dfrac{3.7 \times 2}{\sqrt{3}} = \dfrac{7.4}{1.73} = 4.28 (approx.)

So the ladder should be m long — the slider found it to within a few centimetres.

For the foot of the ladder, DC is adjacent to the 60° angle:

\dfrac{DC}{BD} = \cot 60^\circ = \dfrac{1}{\sqrt{3}}

DC = \dfrac{3.7}{\sqrt{3}} = \dfrac{3.7}{1.73} = 2.14 (approx.)

So she should place the foot of the ladder m from the pole.

Don't Forget the Observer

Example 3 : An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

Solution : In Fig. 9.6, AB is the chimney, CD is the observer and ∠ADE is the angle of elevation. The angle is measured at her eye D, so the right triangle is ADE, right-angled at E: it stops at eye level.

The chimney splits in two: AB = AE + EB, where EB = CD = 1.5 m. Also DE = CB = 28.5 m.

We need a ratio that joins AE and DE, so we take the of the angle of elevation:

\tan 45^\circ = \dfrac{AE}{DE}

1 = \dfrac{AE}{28.5}

So AE = m, and the height of the chimney is AB = (28.5 + 1.5) m = m.

Fig. 9.6 — the chimney AB and the observer CD. (The observer is drawn taller than scale.)

Tip

Forgetting the observer's height is the most common slip in this chapter. The angle is measured from the eye, so the trigonometry only gives the part of the height above eye level. Add the eye height at the end.

The Flag on the Roof

Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take \sqrt{3} = 1.732)

Solution : In Fig. 9.7, AB is the building, BD is the flagstaff and P is the given point. This time there are two right triangles, △PAB and △PAD. We want the length of the flagstaff, DB, and the distance PA.

We know AB, so we start with the right triangle PAB:

\tan 30^\circ = \dfrac{AB}{AP}

\dfrac{1}{\sqrt{3}} = \dfrac{10}{AP}

AP = 10\sqrt{3}

So the building is 10\sqrt{3} m = m from P.

Fig. 9.7 — the building AB, the flagstaff BD and the point P.

Now let DB = x m, so that AD = (10 + x) m. In the right triangle PAD,

\tan 45^\circ = \dfrac{AD}{AP} = \dfrac{10 + x}{10\sqrt{3}}

1 = \dfrac{10 + x}{10\sqrt{3}}

x = 10\sqrt{3} - 10 = 10(\sqrt{3} - 1) = 7.32

So the flagstaff is m long.

Two Shadows, One Tower

Example 5 : The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.

The Sun's altitude is the angle of elevation of the Sun, so it is also the angle of elevation of the top of the tower from the tip of its shadow.

Fig. 9.8 is live. Slide the Sun's altitude and complete both missions.

Sun's altitude: ${sun}°

Solution : In Fig. 9.8, AB is the tower, BC is its shadow when the Sun's altitude is 60°, and BD is its shadow when the altitude is 30°. So ∠ACB = 60° and ∠ADB = 30°.

Let AB = h m and BC = x m. The 30° shadow is 40 m longer, so DB = (x + 40) m. The two right triangles ABC and ABD share the side AB.

In the right triangle ABC,

\tan 60^\circ = \dfrac{AB}{BC}

\sqrt{3} = \dfrac{h}{x}    (1)

In the right triangle ABD,

\tan 30^\circ = \dfrac{AB}{BD}

\dfrac{1}{\sqrt{3}} = \dfrac{h}{x + 40}    (2)

From (1), h = x\sqrt{3}. Putting this value into (2),

(x\sqrt{3})\sqrt{3} = x + 40

3x = x + 40

So x = , and from (1), h = 20\sqrt{3}.

The height of the tower is m — about 34.6 m. The two shadows you marked, 20 m and 60 m, are exactly x and x + 40.

Looking Down from the Top Floor

Example 6 : The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Solution : In Fig. 9.9, PC is the multi-storeyed building and AB is the 8 m building. We want the height PC and the distance AC.

Fig. 9.9 — the multi-storeyed building PC and the 8 m building AB.

Both given angles are angles of depression, measured down from the horizontal PQ. The horizontals PQ and BD are parallel, and PB is a transversal to them. So the angles QPB and PBD are angles, and so they are equal.

So ∠PBD = 30°. In the same way, ∠PAC = 45°. That gives two right triangles to work with: triangle PBD and triangle PAC.

In the right triangle PBD,

\dfrac{PD}{BD} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}

BD = PD\sqrt{3}

In the right triangle PAC,

\dfrac{PC}{AC} = \tan 45^\circ = 1

so PC = .

Also PC = PD + DC, and DC = AB = 8 m, so PD + 8 = AC. And AC = BD, because they are opposite sides of the rectangle ABDC. So

PD + 8 = BD = PD\sqrt{3}

PD(\sqrt{3} - 1) = 8

PD = \dfrac{8}{\sqrt{3} - 1} = \dfrac{8(\sqrt{3} + 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = 4(\sqrt{3} + 1) m

So the height of the multi-storeyed building is PC = PD + 8 = 4(\sqrt{3} + 1) + 8 m = m, and the distance between the two buildings, AC = PC, is also m.

The River Below the Bridge

Example 7 : From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Solution : In Fig. 9.10, A and B are points on opposite banks, so AB is the width of the river. P is the point on the bridge and D is the point directly below it, so DP = 3 m.

Both angles of depression are measured from the horizontal through P. As alternate angles, they reappear as ∠A = 30° and ∠B = 45°. The width splits as AB = AD + DB.

Fig. 9.10 — the point P on the bridge, 3 m above the banks A and B.

In the right triangle APD, ∠A = 30°, so

\tan 30^\circ = \dfrac{PD}{AD}

\dfrac{1}{\sqrt{3}} = \dfrac{3}{AD}

AD = 3\sqrt{3} m

In the right triangle PBD, ∠B = 45°, so BD = = 3 m.

AB = BD + AD = 3 + 3\sqrt{3} = 3(1 + \sqrt{3}) m

So the river is m wide — about 8.2 m.

Speed Round: One Look, One Answer

The last round of this section. Each sighting needs one ratio and one standard value — no calculator, and leave \sqrt{3} as it is.

Get five right to move on.

Sina