What is the angle of elevation and the angle of depression?

The angle of elevation of an object above the horizontal is the angle between the horizontal line of sight and the line joining the eye to the object, formed when you raise your head to look UP at something. The angle of depression is the same idea when the object is BELOW the horizontal and you lower your head to look DOWN at it -- by alternate angles, the angle of depression from a higher point always equals the angle of elevation measured from the lower point back up to the first.

How do you solve a heights and distances problem step by step?

Draw a simple diagram first, marking every given height, distance and angle as a right triangle (or a pair of right triangles sharing a common side). Identify which side is opposite the known angle and which is adjacent, pick the trigonometric ratio that links the side you know with the side you want, then solve the resulting equation.

Why do two-triangle heights and distances questions usually need two equations?

When neither triangle alone contains enough known values, each angle of elevation or depression gives one equation in the two unknowns (commonly the height and the distance). Setting the two expressions for the shared unknown equal to each other, or substituting one equation into the other, then solves for both.

What common values of sin, cos and tan are used in this chapter?

Almost every question in Exercise 9.1 uses only the standard angles 30°, 45° and 60° from Table 8.1 (Chapter 8): tan 30° = 1/root3, tan 45° = 1, tan 60° = root3, and the corresponding sin and cos values, since a 'nice' angle keeps the algebra exact and avoids a calculator.

Does Class 10 Chapter 9 have more than one exercise?

No -- in both the old and the rationalised 2026-27 syllabus, Some Applications of Trigonometry has always been a single-exercise chapter. Exercise 9.1 (15 questions) is followed directly by the chapter Summary, with nothing dropped in the rationalisation.

NCERT Solutions Class 10 Maths Ch 9 Heights and Distances

7 Examples + 15 Questions Solved

Chapter 9: Some Applications of Trigonometry — NCERT Solutions (rationalised syllabus 2026–27), with all 7 worked Examples from the chapter plus every question in Exercise 9.1. This chapter applies the trigonometric ratios from Chapter 8 to real heights-and-distances problems. Two ideas run through every question: the angle of elevation is the angle between the horizontal line of sight and the line joining the eye to an object ABOVE the horizontal (raising your head to look up), and the angle of depression is the same idea for an object BELOW the horizontal (lowering your head to look down). By alternate angles, an angle of depression measured from a higher point always equals the angle of elevation measured back up from the lower point.

Every question reduces to one or two right triangles: draw a simple diagram, mark every given height, distance and angle, identify which side is opposite the known angle and which is adjacent, then pick the ratio (usually tan, sometimes sin or cos) that links the side you know with the side you want. Where a question gives TWO angles from the same point, or the same object seen from two different points, it produces two equations in two unknowns — solved either by substitution or by setting two expressions for the same unknown equal to each other.

Almost every angle in this chapter is one of the standard values 30°, 45° or 60° from Table 8.1 (Chapter 8): \(\tan30^{\circ}=\dfrac{1}{\sqrt3}\), \(\tan45^{\circ}=1\), \(\tan60^{\circ}=\sqrt3\), keeping every answer an exact surd rather than a calculator decimal — confirmed page by page, this single-exercise chapter is unchanged in shape between the old and the rationalised syllabus: Exercise 9.1 (15 questions) is followed directly by the Summary, with nothing dropped.

Worked Examples (Examples 1–7)

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.

Fig. 9.4 — tower AB, point C is 15 m from the foot B, angle of elevation ∠ACB = 60°

Solution

Let AB be the tower and C the point on the ground, 15 m from the foot B, with the angle of elevation ∠ACB = 60° (Fig. 9.4). We need to find AB, and we know the side CB = 15 m — so we pick the ratio that links the side we know (CB, adjacent to ∠C) with the side we want (AB, opposite ∠C): tan C.

\(\tan C=\dfrac{AB}{CB}\)

\(\tan60^{\circ}=\dfrac{AB}{15}\)

\(\sqrt3=\dfrac{AB}{15}\ \Rightarrow\ AB=15\sqrt3\ \text{m}\)

Hence, the height of the tower is \(15\sqrt3\) m (≈ 25.98 m).

Example 2

An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. 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 \(\sqrt3=1.73\))

Fig. 9.5 — pole AD = 5 m, point B is 1.3 m below the top A, ladder BC inclined at 60° to the ground

Solution

In Fig. 9.5, AD is the pole (AD = 5 m) and B is the point the electrician must reach, 1.3 m below the top A. C is the foot of the ladder, and the ladder BC is inclined at 60° to the horizontal ground DC.

First find BD, the height of B above the ground:

\(BD=AD-AB=(5-1.3)\ \text{m}=3.7\ \text{m}\)

Now, BC is the ladder — the hypotenuse of the right \(\triangle\)BDC. We know BD and the angle at C, so we use sin 60° (opposite over hypotenuse):

\(\dfrac{BD}{BC}=\sin60^{\circ}\)

\(\dfrac{3.7}{BC}=\dfrac{\sqrt3}{2}\)

\(BC=\dfrac{3.7\times2}{\sqrt3}=\dfrac{7.4}{1.73}=4.28\ \text{m (approx.)}\)

So the length of the ladder should be 4.28 m.

Next, DC is the distance of the foot of the ladder from the pole. We know BD (opposite ∠C) and want DC (adjacent to ∠C), so we use cot 60°:

\(\dfrac{DC}{BD}=\cot60^{\circ}=\dfrac{1}{\sqrt3}\)

\(DC=\dfrac{3.7}{\sqrt3}=\dfrac{3.7}{1.73}=2.14\ \text{m (approx.)}\)

Therefore, she should place the foot of the ladder at a distance of 2.14 m from the pole.

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?

Fig. 9.6 — chimney AB, observer’s eye at D (1.5 m above the ground C), 28.5 m from the chimney, angle of elevation ∠ADE = 45°

Solution

In Fig. 9.6, AB is the chimney and CD the observer, with \(\angle\)ADE = 45° the angle of elevation (\(\triangle\)ADE is right-angled at E). We are required to find the height of the chimney, AB.

We have

\(AB=AE+BE=AE+1.5\)

and

\(DE=CB=28.5\ \text{m}\)

To find AE, we choose a ratio that involves both AE and DE. Since we know DE and want AE, and DAE is right-angled at E, we use the tangent of the angle of elevation:

\(\tan45^{\circ}=\dfrac{AE}{DE}\)

\(1=\dfrac{AE}{28.5}\)

\(AE=28.5\ \text{m}\)

So the height of the chimney is:

\(AB=(28.5+1.5)\ \text{m}=30\ \text{m}\)

Hence, the height of the chimney is 30 m.

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 \(\sqrt3=1.732\))

Fig. 9.7 — building AB = 10 m with flagstaff BD on top; from P, angle of elevation to B is 30° and to D is 45°

Solution

In Fig. 9.7, AB denotes the height of the building, BD the flagstaff, and P the given point. Note that there are two right triangles PAB and PAD, sharing the base PA. We are required to find the length of the flagstaff, DB, and the distance of the building from P, i.e. PA.

Since we know the height of the building AB, we first consider right \(\triangle\)PAB, using the ratio that links the known side AB with the side PA we want:

\(\tan30^{\circ}=\dfrac{AB}{AP}\)

\(\dfrac{1}{\sqrt3}=\dfrac{10}{AP}\)

\(AP=10\sqrt3=17.32\ \text{m}\)

So the distance of the building from P is \(10\sqrt3\) m = 17.32 m.

Next, let DB = x m, so AD = (10 + x) m. Now, in right \(\triangle\)PAD (angle of elevation 45°):

\(\tan45^{\circ}=\dfrac{AD}{AP}=\dfrac{10+x}{10\sqrt3}\)

\(1=\dfrac{10+x}{10\sqrt3}\)

\(x=10\left(\sqrt3-1\right)\)

\(x=10(1.732-1)=7.32\ \text{m}\)

So, the length of the flagstaff is 7.32 m.

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.

Fig. 9.8 — tower AB, shadow BC when the sun’s altitude is 60°, shadow DB (40 m longer) when the altitude is 30°

Solution

In Fig. 9.8, AB is the tower and BC is the length of the shadow when the Sun’s altitude (the angle of elevation of the top of the tower from the tip of the shadow) is 60°, and DB is the length of the shadow when the altitude is 30°, with DB exactly 40 m longer than BC.

Let AB be \(h\) m and BC be \(x\) m. According to the question, DB is 40 m longer than BC:

\(DB=(40+x)\ \text{m}\)

Now we have two right triangles, ABC and ABD.

In right \(\triangle\)ABC:

\(\tan60^{\circ}=\dfrac{AB}{BC}\)

\(\sqrt3=\dfrac{h}{x}\ \ \text{...(1)}\)

In right \(\triangle\)ABD:

\(\tan30^{\circ}=\dfrac{AB}{BD}\)

\(\dfrac{1}{\sqrt3}=\dfrac{h}{x+40}\ \ \text{...(2)}\)

From (1), \(h=x\sqrt3\). Substituting this into (2):

\(\left(x\sqrt3\right)\sqrt3=x+40\)

\(3x=x+40\)

\(x=20\)

So, from (1), \(h=x\sqrt3=20\sqrt3\).

Therefore, the height of the tower is \(20\sqrt3\) m (≈ 34.64 m).

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.

Fig. 9.9 — multi-storeyed building PC, 8 m building AB; from the horizontal PQ, the angle of depression to the top B is 30° and to the foot A is 45°

Solution

In Fig. 9.9, PC denotes the multi-storeyed building and AB the 8 m tall building. We are required to find the height of the multi-storeyed building (PC) and the distance between the buildings (AC).

PB is a transversal to the parallel lines PQ (horizontal through P) and BD (horizontal through B). So \(\angle\)QPB and \(\angle\)PBD are alternate angles, and therefore equal, so \(\angle\)PBD = 30°. Similarly, \(\angle\)PAC = 45°.

In right \(\triangle\)PBD:

\(\dfrac{PD}{BD}=\tan30^{\circ}=\dfrac{1}{\sqrt3}\ \ \text{or}\ \ BD=PD\sqrt3\)

In right \(\triangle\)PAC:

\(\dfrac{PC}{AC}=\tan45^{\circ}=1\ \Rightarrow\ PC=AC\)

i.e. PC = AC. Also, PC = PD + DC, so PD + DC = AC. Since AC = BD (opposite sides of the rectangle ABDC) and DC = AB = 8 m, we get:

\(PD+8=BD=PD\sqrt3\ \ (\text{since }AC=BD\text{, }DC=AB=8\text{ m})\)

This gives:

\(PD\left(\sqrt3-1\right)=8\)

\(PD=\dfrac{8}{\sqrt3-1}=\dfrac{8(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{8(\sqrt3+1)}{2}=4(\sqrt3+1)\ \text{m}\)

So, the height of the multi-storeyed building is \(\{4(\sqrt3+1)+8\}\) m = \(4(3+\sqrt3)\) m (≈ 18.93 m), and the distance between the two buildings is also \(4(3+\sqrt3)\) m, since it equals PC exactly (both equal AC, and AC = PC directly from the 45° triangle above).

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.

Fig. 9.10 — bridge point P is 3 m above the river; the 30° and 45° angles of depression from P are marked at A and B as the equal angles of elevation

Solution

In Fig. 9.10, A and B represent points on the bank on opposite sides of the river, so AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e. DP = 3 m. We are required to find the width of the river, AB = AD + DB.

In right \(\triangle\)APD, \(\angle\)A = 30° (the angle of depression from P equals the angle of elevation from A, being alternate angles with the horizontal at P).

\(\tan30^{\circ}=\dfrac{PD}{AD}\)

\(\dfrac{1}{\sqrt3}=\dfrac{3}{AD}\)

\(AD=3\sqrt3\ \text{m}\)

Also, in right \(\triangle\)PBD, \(\angle\)B = 45°. So:

\(BD=PD=3\ \text{m (since }\tan45^{\circ}=1\text{)}\)

Now:

\(AB=AD+DB=3\sqrt3+3=3\left(\sqrt3+1\right)\ \text{m}\)

Therefore, the width of the river is \(3(\sqrt3+1)\) m (≈ 8.196 m).

Exercise 9.1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Fig. 9.11 — pole AB, rope AC = 20 m tied from the top to the ground at C, angle with the ground = 30°

Answer

Let AB be the pole and AC the rope (AC = 20 m), tied from the top A to a point C on the ground, with \(\angle\)ACB = 30° and the right angle at B (Fig. 9.11). We want AB, and we know AC (the hypotenuse) and the angle at C, so we use sin C (opposite over hypotenuse):

\(\sin C=\dfrac{AB}{AC}\)

\(\sin30^{\circ}=\dfrac{AB}{20}\)

\(\dfrac12=\dfrac{AB}{20}\ \Rightarrow\ AB=10\ \text{m}\)

Therefore, the height of the pole is 10 m.

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer

Let the tree originally stand at B, breaking at a point A above the ground, with the broken part AC bending over to touch the ground at C, where \(\angle\)ACB = 30°, the right angle at B (foot of the tree), and BC = 8 m.

The original height of the tree is the standing stump AB plus the broken part AC (which now lies along the hypotenuse, but was originally the same length as the upper part of the tree). So we need both AB and AC.

For AB (opposite ∠C, adjacent side BC known), use tan C:

\(\tan C=\dfrac{AB}{BC}\)

\(\tan30^{\circ}=\dfrac{AB}{8}\)

\(AB=\dfrac{8}{\sqrt3}\ \text{m}\)

For AC (hypotenuse, adjacent side BC known), use cos C:

\(\cos C=\dfrac{BC}{AC}\)

\(\cos30^{\circ}=\dfrac{8}{AC}\)

\(AC=\dfrac{8}{\cos30^{\circ}}=\dfrac{8}{\left(\sqrt3/2\right)}=\dfrac{16}{\sqrt3}\ \text{m}\)

So the total height of the tree is:

\(AB+AC=\dfrac{8}{\sqrt3}+\dfrac{16}{\sqrt3}=\dfrac{24}{\sqrt3}\)

\(=\dfrac{24\sqrt3}{3}=8\sqrt3\ \text{m}\)

Therefore, the height of the tree is \(8\sqrt3\) m (≈ 13.86 m).

A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer

Each slide is the hypotenuse of a right triangle whose height (opposite side) and base angle are given, so in both cases we use the sine ratio: \(\sin(\text{angle})=\dfrac{\text{height}}{\text{length of slide}}\).

Slide for younger children (height 1.5 m, angle 30°):

\(\sin30^{\circ}=\dfrac{1.5}{\text{length}}\)

\(\dfrac12=\dfrac{1.5}{\text{length}}\)

\(\text{length}=3\ \text{m}\)

Slide for elder children (height 3 m, angle 60°):

\(\sin60^{\circ}=\dfrac{3}{\text{length}}\)

\(\dfrac{\sqrt3}{2}=\dfrac{3}{\text{length}}\)

\(\text{length}=\dfrac{6}{\sqrt3}=2\sqrt3\ \text{m}\)

So the slide for the younger children should be 3 m long, and the steeper slide for the elder children should be \(2\sqrt3\) m (≈ 3.46 m) long.

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer

Let AB be the tower and C the point on the ground, 30 m from the foot B, with \(\angle\)ACB = 30°. We know CB and want AB, so we use tan C:

\(\tan C=\dfrac{AB}{CB}\)

\(\tan30^{\circ}=\dfrac{AB}{30}\)

\(AB=30\times\dfrac{1}{\sqrt3}=\dfrac{30}{\sqrt3}=10\sqrt3\ \text{m}\)

Therefore, the height of the tower is \(10\sqrt3\) m (≈ 17.32 m).

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer

The kite's height (60 m) is the side opposite the 60° angle of inclination, and the string is the hypotenuse. We know the opposite side and want the hypotenuse, so we use sin 60°:

\(\sin60^{\circ}=\dfrac{\text{height}}{\text{length of string}}\)

\(\dfrac{\sqrt3}{2}=\dfrac{60}{\text{length}}\)

\(\text{length}=\dfrac{120}{\sqrt3}=40\sqrt3\ \text{m}\)

Therefore, the length of the string is \(40\sqrt3\) m (≈ 69.28 m).

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Only the part of the building ABOVE the boy's eye level matters for the angle of elevation, so the effective height is:

\(\text{effective height}=30-1.5=28.5\ \text{m}\)

Let \(d_1\) be his distance from the building at the FARTHER position (angle 30°) and \(d_2\) at the NEARER position (angle 60°). In each case, the effective height (28.5 m) is opposite the angle, and the distance is adjacent, so we use the tangent ratio.

At the farther point:

\(\tan30^{\circ}=\dfrac{28.5}{d_1}\)

\(d_1=\dfrac{28.5}{\left(1/\sqrt3\right)}=28.5\sqrt3\ \text{m}\)

At the nearer point:

\(\tan60^{\circ}=\dfrac{28.5}{d_2}\)

\(d_2=\dfrac{28.5}{\sqrt3}\ \text{m}\)

The distance he walked towards the building is \(d_1-d_2\):

\(d_1-d_2=28.5\sqrt3-\dfrac{28.5}{\sqrt3}=28.5\left(\sqrt3-\dfrac{1}{\sqrt3}\right)\)

\(=28.5\times\dfrac{3-1}{\sqrt3}=\dfrac{57}{\sqrt3}\)

\(=\dfrac{57\sqrt3}{3}=19\sqrt3\ \text{m}\)

Therefore, he walked a distance of \(19\sqrt3\) m (≈ 32.91 m) towards the building.

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer

Let the point on the ground be P, the building AB = 20 m (bottom of the tower is at A, the top of the building), and the tower AC on top, so we want CA, the height of the tower.

Since the angle of elevation to the bottom of the tower (A, top of the building) is 45°, and we know AB = 20 m:

\(\tan45^{\circ}=\dfrac{AB}{PB}\)

\(1=\dfrac{20}{PB}\)

\(PB=20\ \text{m}\)

Now, using the angle of elevation to the top of the tower (60°), for the TOTAL height (building + tower):

\(\tan60^{\circ}=\dfrac{AB+CA}{PB}\)

\(\sqrt3=\dfrac{20+CA}{20}\)

\(20+CA=20\sqrt3\)

The height of the tower alone is:

\(CA=20\sqrt3-20\)

\(CA=20\left(\sqrt3-1\right)\ \text{m}\)

Therefore, the height of the tower is \(20(\sqrt3-1)\) m (≈ 14.64 m).

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer

Let the pedestal height be h m and the horizontal distance from the observation point be d m. Using the angle of elevation of the top of the pedestal (45°):

\(\tan45^{\circ}=\dfrac{h}{d}\)

\(1=\dfrac{h}{d}\ \Rightarrow\ d=h\)

Using the angle of elevation of the top of the statue (60°), the total height is (h + 1.6):

\(\tan60^{\circ}=\dfrac{h+1.6}{d}\)

\(\sqrt3=\dfrac{h+1.6}{d}\)

Substituting d = h:

\(\sqrt3=\dfrac{h+1.6}{h}\)

\(h\sqrt3-h=1.6\)

\(h(\sqrt3-1)=1.6\)

\(h=\dfrac{1.6}{\sqrt3-1}=\dfrac{1.6(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{1.6(\sqrt3+1)}{2}=0.8(\sqrt3+1)\ \text{m}\)

Therefore, the height of the pedestal is \(0.8(\sqrt3+1)\) m (≈ 2.19 m).

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer

Let the height of the building be h m and the distance between the tower and the building be d m. Using the angle of elevation of the top of the TOWER (50 m) from the foot of the building (60°):

\(\tan60^{\circ}=\dfrac{50}{d}\)

\(\sqrt3=\dfrac{50}{d}\)

\(d=\dfrac{50}{\sqrt3}\ \text{m}\)

Using the angle of elevation of the top of the BUILDING from the foot of the tower (30°):

\(\tan30^{\circ}=\dfrac{h}{d}\)

\(\dfrac{1}{\sqrt3}=\dfrac{h}{50/\sqrt3}\)

\(h=\dfrac{50}{\sqrt3}\times\dfrac{1}{\sqrt3}=\dfrac{50}{3}\ \text{m}\)

Therefore, the height of the building is \(\dfrac{50}{3}\) m (≈ 16.67 m).

Q10

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer

Let the height of each pole be h m, and let the point on the road be at a distance x m from the pole seen at 60°, so it is (80 − x) m from the pole seen at 30°.

\(\tan60^{\circ}=\dfrac{h}{x}\ \Rightarrow\ h=x\sqrt3\ \ \text{...(1)}\)

\(\tan30^{\circ}=\dfrac{h}{80-x}\)

Rearranging this to make h the subject:

\(h=\dfrac{80-x}{\sqrt3}\ \ \text{...(2)}\)

Setting the two expressions for h equal:

\(x\sqrt3=\dfrac{80-x}{\sqrt3}\)

\(3x=80-x\ \Rightarrow\ 4x=80\)

\(x=20\)

So x = 20 m, and the distance from the other pole is:

\(80-x=80-20=60\ \text{m}\)

The height of each pole is:

\(h=x\sqrt3=20\sqrt3\ \text{[From (1)]}\)

Therefore, the height of each pole is \(20\sqrt3\) m (≈ 34.64 m), and the point is 20 m from the pole seen at 60° and 60 m from the pole seen at 30°.

Q11

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Fig. 9.12 — TV tower AB on a canal bank; from C (directly opposite) angle of elevation 60°, from D (20 m further back) angle of elevation 30°

Answer

In Fig. 9.12, let AB be the TV tower and C the point on the opposite bank, directly across the canal, so CB is the width of the canal. D is a further point, 20 m from C, on the line CB extended, with \(\angle\)ACB = 60° and \(\angle\)ADB = 30°. Let CB = x m and AB = h m.

\(\tan60^{\circ}=\dfrac{h}{x}\ \Rightarrow\ h=x\sqrt3\ \ \text{...(1)}\)

\(\tan30^{\circ}=\dfrac{h}{x+20}\)

\(h=\dfrac{x+20}{\sqrt3}\ \ \text{...(2)}\)

Setting the two expressions for h equal:

\(x\sqrt3=\dfrac{x+20}{\sqrt3}\)

\(3x=x+20\)

\(2x=20\)

\(x=10\)

So the width of the canal is x = 10 m, and the height of the tower is:

\(h=x\sqrt3=10\sqrt3\ \text{[From (1)]}\)

Therefore, the height of the tower is \(10\sqrt3\) m (≈ 17.32 m), and the width of the canal is 10 m.

Q12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer

Let the top of the 7 m building be P, and the cable tower stand with its foot at the ground level. Since the angle of depression from P to the foot of the tower is 45°, that equals the angle of elevation of P from the foot of the tower (alternate angles). Let the horizontal distance between the building and the tower be d m:

\(\tan45^{\circ}=\dfrac{7}{d}\)

\(1=\dfrac{7}{d}\)

\(d=7\ \text{m}\)

Now, the angle of elevation from P to the TOP of the cable tower is 60°. The extra height above the level of P (call it e) satisfies:

\(\tan60^{\circ}=\dfrac{e}{d}\)

\(\sqrt3=\dfrac{e}{7}\)

\(e=7\sqrt3\ \text{m}\)

The total height of the cable tower is the 7 m up to P’s level, plus the extra height e above it:

\(\text{height of tower}=7+7\sqrt3\)

\(=7\left(1+\sqrt3\right)\ \text{m}\)

Therefore, the height of the tower is \(7(1+\sqrt3)\) m (≈ 19.12 m).

Q13

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer

Let the foot of the lighthouse be O, its top be P (OP = 75 m), and let the nearer ship (angle of depression 45°) be at a distance \(d_1\), the farther ship (angle of depression 30°) at a distance \(d_2\) — both angles of depression equal the corresponding angles of elevation from the ships, by alternate angles.

For the nearer ship:

\(\tan45^{\circ}=\dfrac{75}{d_1}\)

\(1=\dfrac{75}{d_1}\)

\(d_1=75\ \text{m}\)

For the farther ship:

\(\tan30^{\circ}=\dfrac{75}{d_2}\)

\(\dfrac{1}{\sqrt3}=\dfrac{75}{d_2}\)

\(d_2=75\sqrt3\ \text{m}\)

The distance between the two ships is:

\(d_2-d_1=75\sqrt3-75\)

\(=75\left(\sqrt3-1\right)\ \text{m}\)

Therefore, the distance between the two ships is \(75(\sqrt3-1)\) m (≈ 54.9 m).

Q14

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Fig. 9.13 — girl 1.2 m tall, balloon moving along a horizontal line 88.2 m up; angle of elevation reduces from 60° to 30°

Answer

Only the part of the balloon's height ABOVE the girl's eye level matters for the angle of elevation, so the effective height is:

\(\text{effective height}=88.2-1.2=87\ \text{m}\)

Let \(d_1\) be the horizontal distance to the balloon at the FIRST instant (angle 60°), and \(d_2\) at the LATER instant (angle 30°), both measured from a point directly below the girl's eyes.

\(\tan60^{\circ}=\dfrac{87}{d_1}\ \Rightarrow\ d_1=\dfrac{87}{\sqrt3}\ \text{m}\)

\(\tan30^{\circ}=\dfrac{87}{d_2}\ \Rightarrow\ d_2=87\sqrt3\ \text{m}\)

Since the balloon travels horizontally, the distance it travels is \(d_2-d_1\):

\(d_2-d_1=87\sqrt3-\dfrac{87}{\sqrt3}\)

\(=87\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=87\times\dfrac{3-1}{\sqrt3}=\dfrac{174}{\sqrt3}\)

\(=\dfrac{174\sqrt3}{3}=58\sqrt3\ \text{m}\)

Therefore, the distance travelled by the balloon is \(58\sqrt3\) m (≈ 100.46 m).

Q15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer

Let the height of the tower be h m. At the FIRST sighting (angle of depression, hence angle of elevation from the car, 30°), let the car be at distance \(d_1\) from the foot of the tower; six seconds later (angle 60°), at distance \(d_2\).

\(\tan30^{\circ}=\dfrac{h}{d_1}\ \Rightarrow\ d_1=h\sqrt3\)

\(\tan60^{\circ}=\dfrac{h}{d_2}\ \Rightarrow\ d_2=\dfrac{h}{\sqrt3}\)

The distance covered by the car in those 6 seconds is:

\(d_1-d_2=h\sqrt3-\dfrac{h}{\sqrt3}\)

\(=h\left(\sqrt3-\dfrac{1}{\sqrt3}\right)=h\times\dfrac{3-1}{\sqrt3}=\dfrac{2h}{\sqrt3}\)

\(\text{distance covered in 6 s}=\dfrac{2h}{\sqrt3}\ \text{m}\)

So the car's (uniform) speed is:

\(\text{speed}=\dfrac{2h/\sqrt3}{6}=\dfrac{h}{3\sqrt3}\ \text{m/s}\)

We want the further time to cover the REMAINING distance, \(d_2\) (from the second sighting point to the foot of the tower):

\(\text{time}=\dfrac{d_2}{\text{speed}}=\dfrac{h/\sqrt3}{h/(3\sqrt3)}\)

\(=\dfrac{h}{\sqrt3}\times\dfrac{3\sqrt3}{h}\)

\(=3\ \text{s}\)

Therefore, the car takes a further 3 seconds to reach the foot of the tower.

Chapter 8: Introduction to Trigonometry

All Chapters

Chapter 10: Circles

Some Applications of Trigonometry

Worked Examples (Examples 1–7)

Exercise 9.1

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