NCERT Class 10 Mathematics — Some Applications of Trigonometry, Exercise 9.1. All 15 questions solved.

Exercise 9.1 is the whole of Chapter 9: fifteen word problems, each of which turns into one or two right triangles once the figure is drawn. Three terms carry the wording:

• The line of sight is the line from the observer’s eye to the object.
• The angle of elevation is the angle between the line of sight and the horizontal when the object is above eye level.
• The angle of depression is the same angle when the object is below eye level — measured down from the horizontal, never from the vertical.

The ratios needed are the standard ones: $\tan 30^\circ = \frac{1}{\sqrt3}$, $\tan 45^\circ = 1$, $\tan 60^\circ = \sqrt3$, $\sin 30^\circ = \frac12$ and $\sin 60^\circ = \frac{\sqrt3}{2}$. Use tangent when the two lengths involved are the height and the horizontal distance, and sine when one of them is a slanting length — a rope, a string, a slide.

Key insight. In every two-angle problem the two right triangles share a side, usually the vertical height or the horizontal distance. Write the tangent of each angle in terms of that shared side and the unknown, and the unknown drops out of a pair of simple equations. And an angle of depression from the top of a tower equals the angle of elevation of the top from the object below, because the two are alternate angles between parallel horizontal lines — so questions 12, 13 and 15 become elevation problems as soon as the angle is moved to the ground.

Decimal values below use $\sqrt3 \approx 1.732$.

Question 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 $30^\circ$ (see Fig. 9.11).

The pole is $AB$, standing vertically at $B$. The rope runs from the top $A$ to the point $C$ on the ground, so $AC = 20$ m, $\angle ACB = 30^\circ$ and $\angle ABC = 90^\circ$.

Solution. The rope is the hypotenuse of the right triangle $ABC$, and the height $AB$ is the side opposite the $30^\circ$ angle. Opposite and hypotenuse together mean sine:

$$\sin 30^\circ = \frac{AB}{AC} \quad\Longrightarrow\quad \frac12 = \frac{AB}{20} \quad\Longrightarrow\quad AB = 10 \text{ m}$$

Question 2

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^\circ$ 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.

The tree stood at $B$ and snapped at $A$. The standing part $AB$ is still vertical; the broken part $AC$, still joined at $A$, leans over so that the old top $C$ rests on the ground with $BC = 8$ m and $\angle ACB = 30^\circ$. The original height of the tree is $AB + AC$.

Solution. $BC$ is the side adjacent to the $30^\circ$ angle, so it gives both of the other sides of the triangle.

The standing part $AB$ is opposite the angle, so use tangent:

$$\tan 30^\circ = \frac{AB}{BC} \quad\Longrightarrow\quad AB = \frac{8}{\sqrt3} \text{ m}$$

The broken part $AC$ is the hypotenuse, so use cosine:

$$\cos 30^\circ = \frac{BC}{AC} \quad\Longrightarrow\quad \frac{\sqrt3}{2} = \frac{8}{AC} \quad\Longrightarrow\quad AC = \frac{16}{\sqrt3} \text{ m}$$

Before it broke, the tree was both pieces end to end:

$$AB + AC = \frac{8}{\sqrt3} + \frac{16}{\sqrt3} = \frac{24}{\sqrt3} = \frac{24\sqrt3}{3} = 8\sqrt3 \text{ m} \approx 13.86 \text{ m}$$

Question 3

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 $30^\circ$ 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 $60^\circ$ to the ground. What should be the length of the slide in each case?

Each slide is the hypotenuse of a right triangle whose vertical side is the height of its top: $1.5$ m with the slide at $30^\circ$ to the ground, and $3$ m with the slide at $60^\circ$.

Solution. The length wanted is the hypotenuse and the height is opposite the angle at the ground, so sine connects them. Call the lengths $l_1$ and $l_2$.

For the smaller slide:

$$\sin 30^\circ = \frac{1.5}{l_1} \quad\Longrightarrow\quad l_1 = \frac{1.5}{\frac12} = 3 \text{ m}$$

For the steeper slide:

$$\sin 60^\circ = \frac{3}{l_2} \quad\Longrightarrow\quad l_2 = 3 \cdot \frac{2}{\sqrt3} = \frac{6}{\sqrt3} = 2\sqrt3 \text{ m} \approx 3.46 \text{ m}$$

Doubling the height and steepening the angle almost cancel: the steep slide is only about half a metre longer.

Question 4

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 $30^\circ$. Find the height of the tower.

The tower is $AB$ with foot $B$. The point $C$ is on the ground with $BC = 30$ m, and $\angle ACB = 30^\circ$.

Solution. The height $AB$ is opposite the $30^\circ$ angle and the distance $BC$ is adjacent, so tangent:

$$\tan 30^\circ = \frac{AB}{BC} \quad\Longrightarrow\quad \frac{1}{\sqrt3} = \frac{AB}{30} \quad\Longrightarrow\quad AB = \frac{30}{\sqrt3} = 10\sqrt3 \text{ m} \approx 17.32 \text{ m}$$

Question 5

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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.

The kite is at $A$, directly above the point $B$ on the ground, with $AB = 60$ m. The string is tied at $C$ on the ground and, having no slack, runs straight from $C$ to $A$ with $\angle ACB = 60^\circ$.

Solution. “No slack” is what makes the string a straight line, and so the hypotenuse of the right triangle $ABC$. The height is opposite the $60^\circ$ angle:

$$\sin 60^\circ = \frac{AB}{AC} \quad\Longrightarrow\quad \frac{\sqrt3}{2} = \frac{60}{AC} \quad\Longrightarrow\quad AC = \frac{120}{\sqrt3} = 40\sqrt3 \text{ m} \approx 69.28 \text{ m}$$

Question 6

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 $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.

The building is $AB$, 30 m tall, with foot $B$. The boy’s eyes are 1.5 m above the ground, and the horizontal line through them meets the building at $E$, so $BE = 1.5$ m. From his first position $P$ the angle of elevation of $A$ is $30^\circ$; he walks to $Q$, where it is $60^\circ$. The distance walked is $PQ$.

Solution. The angles are measured at his eyes, so the vertical side of each triangle is not the full 30 m but the part of the building above eye level:

$$AE = AB – BE = 30 – 1.5 = 28.5 \text{ m}$$

In the right triangle $AEP$, the $30^\circ$ angle gives the first distance:

$$\tan 30^\circ = \frac{AE}{PE} \quad\Longrightarrow\quad PE = 28.5\sqrt3 \text{ m}$$

In the right triangle $AEQ$, the $60^\circ$ angle gives the second:

$$\tan 60^\circ = \frac{AE}{QE} \quad\Longrightarrow\quad QE = \frac{28.5}{\sqrt3} \text{ m}$$

The distance walked is the difference:

$$PQ = PE – QE = 28.5\sqrt3 – \frac{28.5}{\sqrt3} = 28.5 \cdot \frac{3 – 1}{\sqrt3} = \frac{57}{\sqrt3} = 19\sqrt3 \text{ m} \approx 32.91 \text{ m}$$

Question 7

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 $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.

The building is $BC$, 20 m high, with foot $B$; the tower $CD$ stands on top of it. From the point $A$ on the ground, $\angle CAB = 45^\circ$ (the bottom of the tower) and $\angle DAB = 60^\circ$ (the top).

Solution. Both triangles $ABC$ and $ABD$ have the ground distance $AB$ as their base, and the $45^\circ$ triangle, whose height is known, gives it:

$$\tan 45^\circ = \frac{BC}{AB} \quad\Longrightarrow\quad 1 = \frac{20}{AB} \quad\Longrightarrow\quad AB = 20 \text{ m}$$

The $60^\circ$ triangle then gives the height of the top of the tower above the ground:

$$\tan 60^\circ = \frac{BD}{AB} \quad\Longrightarrow\quad BD = 20\sqrt3 \text{ m}$$

$BD$ includes the building, so subtract it:

$$CD = BD – BC = 20\sqrt3 – 20 = 20(\sqrt3 – 1) \text{ m} \approx 14.64 \text{ m}$$

Question 8

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 $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

The pedestal is $BC$, of height $h$ metres, with foot $B$; the statue $CD$, 1.6 m tall, stands on it. From the point $A$ on the ground, $\angle CAB = 45^\circ$ and $\angle DAB = 60^\circ$.

Solution. This is question 7 turned round: now the top part is known and the bottom part is not. The shared side is again $AB$.

From the $45^\circ$ triangle:

$$\tan 45^\circ = \frac{BC}{AB} \quad\Longrightarrow\quad AB = BC = h$$

From the $60^\circ$ triangle, with $BD = h + 1.6$:

$$\tan 60^\circ = \frac{BD}{AB} \quad\Longrightarrow\quad \sqrt3 = \frac{h + 1.6}{h} \quad\Longrightarrow\quad \sqrt3\,h = h + 1.6$$

$$h(\sqrt3 – 1) = 1.6 \quad\Longrightarrow\quad h = \frac{1.6}{\sqrt3 – 1}$$

Rationalise by multiplying top and bottom by $\sqrt3 + 1$, using $(\sqrt3 – 1)(\sqrt3 + 1) = 2$:

$$h = \frac{1.6(\sqrt3 + 1)}{2} = 0.8(\sqrt3 + 1) \text{ m} \approx 2.19 \text{ m}$$

Question 9

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

The tower is $AB$, 50 m high, with foot $B$; the building is $CD$, of height $h$, with foot $C$, on the same level ground. From $B$, the top $D$ of the building is at $30^\circ$; from $C$, the top $A$ of the tower is at $60^\circ$.

Solution. The two triangles $ABC$ and $DCB$ share only the ground distance $BC$. Find it from the triangle whose height is known — the tower’s, with its $60^\circ$ angle at $C$:

$$\tan 60^\circ = \frac{AB}{BC} \quad\Longrightarrow\quad BC = \frac{50}{\sqrt3} \text{ m}$$

Now the building’s triangle, with its $30^\circ$ angle at $B$:

$$\tan 30^\circ = \frac{CD}{BC} \quad\Longrightarrow\quad h = \frac{BC}{\sqrt3} = \frac{50}{\sqrt3 \cdot \sqrt3} = \frac{50}{3} = 16\tfrac23 \text{ m}$$

A sanity check: the building is seen at the smaller angle across the same distance, so it must be the shorter of the two — and $16\frac23 < 50$.

Question 10

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 $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.

The poles are $AB$ and $CD$, each of height $h$, with feet $B$ and $D$ on opposite sides of the road, so $BD = 80$ m. The point $P$ lies on $BD$ with $BP = x$ and $PD = 80 – x$. From $P$, the top $A$ is at $60^\circ$ and the top $C$ is at $30^\circ$.

Solution. The two triangles share the height $h$, since the poles are equal. Express $h$ from each:

$$\text{In } \triangle ABP:\ \tan 60^\circ = \frac{h}{x} \quad\Longrightarrow\quad h = \sqrt3\,x$$

$$\text{In } \triangle CDP:\ \tan 30^\circ = \frac{h}{80 – x} \quad\Longrightarrow\quad h = \frac{80 – x}{\sqrt3}$$

Equate the two, and multiply through by $\sqrt3$:

$$\sqrt3\,x = \frac{80 – x}{\sqrt3} \quad\Longrightarrow\quad 3x = 80 – x \quad\Longrightarrow\quad x = 20$$

So $h = 20\sqrt3$ m $\approx 34.64$ m. The point is $20$ m from the pole whose top is seen at $60^\circ$ and $80 – 20 = 60$ m from the other. That is what should happen: the steeper angle belongs to the nearer pole.

Question 11

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 $60^\circ$. 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 $30^\circ$ (see Fig. 9.12). Find the height of the tower and the width of the canal.

The tower is $AB$ with foot $B$ on one bank. $C$ is on the opposite bank directly across from $B$, so $BC$ is the width of the canal. $D$ is 20 m further back from $C$, on the line $BC$ extended away from the tower. $\angle ACB = 60^\circ$ and $\angle ADB = 30^\circ$.

Solution. Let the width $BC = x$ m. The two triangles $ABC$ and $ABD$ share the height $AB$, so express it from each:

$$\tan 60^\circ = \frac{AB}{x} \quad\Longrightarrow\quad AB = \sqrt3\,x$$

$$\tan 30^\circ = \frac{AB}{x + 20} \quad\Longrightarrow\quad AB = \frac{x + 20}{\sqrt3}$$

Equating, $\sqrt3\,x = \dfrac{x + 20}{\sqrt3}$, so $3x = x + 20$ and $x = 10$. Then

$$AB = 10\sqrt3 \text{ m} \approx 17.32 \text{ m}$$

Question 12

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

The building is $AB$, 7 m high, with top $A$ and foot $B$. The cable tower is $CD$, with foot $C$ and top $D$, on the same level ground. The horizontal line through $A$ meets the tower at $E$, so $CE = AB = 7$ m and $AE = BC$. From $A$, the top $D$ is at an angle of elevation of $60^\circ$ and the foot $C$ at an angle of depression of $45^\circ$.

Solution. The horizontal line $AE$ splits the tower into a part below eye level, $CE$, which is known, and a part above it, $DE$, which is not. First find the horizontal distance $AE$ from the lower triangle.

The angle of depression $\angle EAC = 45^\circ$, and in the right triangle $AEC$ the side opposite it is $CE = 7$ m:

$$\tan 45^\circ = \frac{CE}{AE} \quad\Longrightarrow\quad AE = 7 \text{ m}$$

Now the upper triangle $AED$, with its $60^\circ$ angle of elevation at $A$:

$$\tan 60^\circ = \frac{DE}{AE} \quad\Longrightarrow\quad DE = 7\sqrt3 \text{ m}$$

The tower is both parts together:

$$CD = CE + DE = 7 + 7\sqrt3 = 7(\sqrt3 + 1) \text{ m} \approx 19.12 \text{ m}$$

Question 13

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

The lighthouse is $AB$, 75 m above sea level, with top $A$ and foot $B$. The ships $C$ and $D$ are in line with $B$ on the same side, $D$ behind $C$. From $A$ the angle of depression of $C$ is $45^\circ$ and of $D$ is $30^\circ$ — the nearer ship is seen at the steeper angle. By alternate angles, $\angle ACB = 45^\circ$ and $\angle ADB = 30^\circ$.

Solution. Moving each angle of depression down to its ship turns both into ordinary right triangles on the base line $BD$, sharing the height $AB = 75$ m.

$$\text{In } \triangle ABC:\ \tan 45^\circ = \frac{AB}{BC} \quad\Longrightarrow\quad BC = 75 \text{ m}$$

$$\text{In } \triangle ABD:\ \tan 30^\circ = \frac{AB}{BD} \quad\Longrightarrow\quad BD = 75\sqrt3 \text{ m}$$

The ships are separated by the difference:

$$CD = BD – BC = 75\sqrt3 – 75 = 75(\sqrt3 – 1) \text{ m} \approx 54.9 \text{ m}$$

Question 14

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 $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

The girl’s eyes are at $A$, 1.2 m above the ground. The balloon is first at $P$ and later at $Q$, both 88.2 m above the ground on the same horizontal line. $B$ and $C$ are the points on the horizontal through $A$ directly below $P$ and $Q$. Then $\angle PAB = 60^\circ$, $\angle QAC = 30^\circ$, and the distance travelled is $PQ = BC$.

Solution. As in question 6, the angles are measured from the girl’s eyes, so the vertical side of each triangle is the balloon’s height above eye level:

$$PB = QC = 88.2 – 1.2 = 87 \text{ m}$$

$$\text{In } \triangle ABP:\ \tan 60^\circ = \frac{PB}{AB} \quad\Longrightarrow\quad AB = \frac{87}{\sqrt3} = 29\sqrt3 \text{ m}$$

$$\text{In } \triangle ACQ:\ \tan 30^\circ = \frac{QC}{AC} \quad\Longrightarrow\quad AC = 87\sqrt3 \text{ m}$$

The balloon moves horizontally, so the distance it travels equals $BC$:

$$PQ = BC = AC – AB = 87\sqrt3 – 29\sqrt3 = 58\sqrt3 \text{ m} \approx 100.46 \text{ m}$$

Question 15

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 $30^\circ$, 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 $60^\circ$. Find the time taken by the car to reach the foot of the tower from this point.

The tower is $AB$, of height $h$, with top $A$ and foot $B$. The car is first at $D$, seen from $A$ at an angle of depression of $30^\circ$, and six seconds later at $C$, seen at $60^\circ$. By alternate angles, $\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$. The time wanted is for the stretch $CB$.

Solution. Neither the height of the tower nor the speed of the car is given, and neither is needed: at uniform speed, time is proportional to distance, so it is enough to compare $DC$ with $CB$. Express both in terms of $h$.

$$\text{In } \triangle ABD:\ \tan 30^\circ = \frac{h}{BD} \quad\Longrightarrow\quad BD = \sqrt3\,h$$

$$\text{In } \triangle ABC:\ \tan 60^\circ = \frac{h}{BC} \quad\Longrightarrow\quad BC = \frac{h}{\sqrt3}$$

The stretch covered in the six seconds is

$$DC = BD – BC = \sqrt3\,h – \frac{h}{\sqrt3} = \frac{3h – h}{\sqrt3} = \frac{2h}{\sqrt3} = 2 \cdot BC$$

So the remaining distance $CB$ is exactly half of what the car covered in six seconds, and at the same uniform speed it takes half the time:

$$\text{time for } CB = \frac12 \times 6 = 3 \text{ seconds}$$

Common mistakes

• Questions 1, 3 and 5, reaching for tangent by habit. The rope, the slides and the kite string are all hypotenuses, and tangent links only the two legs. With a slanting length given or wanted, the ratio is sine (or cosine).
• Question 2, reporting one piece of the tree. $AB = \frac{8}{\sqrt3}$ m is only the part still standing. The tree’s height is $AB + AC$, because the broken piece used to stand on top of it.
• Questions 6 and 14, forgetting the observer’s height. The angles are measured from the eyes, so the vertical sides are $30 – 1.5 = 28.5$ m and $88.2 – 1.2 = 87$ m. Using the full heights gives $20\sqrt3$ m and $58.8\sqrt3$ m, both wrong.
• Questions 7 and 8, stopping at the combined height. The $60^\circ$ triangle measures from the ground to the top of the tower or statue. The building (in 7) or the statue (in 8) still has to be taken off, or set up as part of the equation.
• Question 9, pairing each angle with the wrong object. Each angle belongs to the triangle containing the top it looks at: the $30^\circ$ angle looks at the building’s top, so it goes with the building’s height, even though it is measured at the tower’s foot.
• Questions 12, 13 and 15, measuring the angle of depression from the vertical. It is measured down from the horizontal through the observer. Drawn against the tower instead, a $30^\circ$ depression becomes a $60^\circ$ angle in the working, and every tangent used turns into its reciprocal. Transfer it to the ground by alternate angles and the problem becomes an elevation problem.
• Question 10, assuming the point is the middle of the road. Equal poles do not mean equal distances; the different angles already show that the point is nearer one pole.
• Question 15, trying to find the speed. The data cannot give it — the tower’s height is unknown — and the question does not need it. The answer comes from the ratio $DC : CB = 2 : 1$.

Practise next

• Chapter 8, Exercise 8.2 — the standard values of $30^\circ$, $45^\circ$ and $60^\circ$ that every question here depends on.
• Chapter 8, Exercise 8.1 — ratios read off a right triangle, which is all a heights-and-distances figure is once it is drawn.