NCERT Class 10 Exercise 9.1 Solutions: Heights and Distances

Val­ues used in this exer­cise#

$$\displaystyle \begin{aligned} \sin 30^\circ &= \tfrac12, & \cos 30^\circ &= \tfrac{\sqrt3}{2}, & \tan 30^\circ &= \tfrac{1}{\sqrt3} \\ \sin 45^\circ &= \tfrac{1}{\sqrt2}, & \cos 45^\circ &= \tfrac{1}{\sqrt2}, & \tan 45^\circ &= 1 \\ \sin 60^\circ &= \tfrac{\sqrt3}{2}, & \cos 60^\circ &= \tfrac12, & \tan 60^\circ &= \sqrt3 \end{aligned}$$

Steps for every prob­lem#

1. Draw a rough fig­ure and mark the right angle, the given angle, the known length and the unknown length.
2. If the observer's height is given, mea­sure the angle from eye level and sub­tract the eye height from the objec­t's height.
3. Choose the ratio that links the known side and the unknown side: $\tan$ for oppo­site and adja­cent sides, $\sin$ for oppo­site side and hypotenuse, $\cos$ for adja­cent side and hypotenuse.
4. Solve, sim­plify the surd, and check the answer by sub­sti­tut­ing it back.

Solu­tions to Exer­cise 9.1#

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 is $30^\circ$.

Fig­ure. Let $AB$ be the pole, with $B$ on the ground, and let $AC = 20$ m be the rope, with $\angle ACB = 30^\circ$ and $\angle ABC = 90^\circ$.

Solu­tion. We know the hypotenuse $AC$ and want the side $AB$ oppo­site the $30^\circ$ angle, so we use sine:

$$\displaystyle \sin 30^\circ = \frac{AB}{AC} \implies AB = 20 \times \frac12 = 10.$$

Check: the base is $BC = 20\cos 30^\circ = 10\sqrt3 \approx 17.32$ m, and $10^2 + (10\sqrt3)^2 = 100 + 300 = 400 = 20^2$.

Answer: the pole is 10 m high.

Ques­tion 2#

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

Fig­ure. Let $B$ be the foot of the tree and $A$ the point where it broke, so $AB$ is still stand­ing. Let $C$ be the point where the top touches the ground. Then $BC = 8$ m, $\angle ACB = 30^\circ$, and the bro­ken part $AC$ is the hypotenuse.

Step 1. Find the stand­ing part:

$$\displaystyle \tan 30^\circ = \frac{AB}{BC} \implies AB = \frac{8}{\sqrt3} = \frac{8\sqrt3}{3}.$$

Step 2. Find the bro­ken part:

$$\displaystyle \cos 30^\circ = \frac{BC}{AC} \implies AC = \frac{8}{\sqrt3/2} = \frac{16}{\sqrt3} = \frac{16\sqrt3}{3}.$$

Step 3. Before the storm, the tree was the stand­ing part plus the bro­ken part:

$$\displaystyle AB + AC = \frac{8}{\sqrt3} + \frac{16}{\sqrt3} = \frac{24}{\sqrt3} = 8\sqrt3.$$

Check: $AB \approx 4.62$ m and $AC \approx 9.24$ m, which add up to $13.86$ m. Also $\displaystyle AB^2 + 8^2 = \frac{64}{3} + 64 = \frac{256}{3} = AC^2$.

Answer: the height of the tree is $8\sqrt3$ m $\approx 13.86$ m.

Why this works: the tree's orig­i­nal height is the sum of both pieces, not just the part still stand­ing.

Ques­tion 3#

A con­trac­tor plans to install two slides in a park. For chil­dren below 5 years, the top of the slide is at a height of 1.5 m and the slide is inclined at $30^\circ$ to the ground. For elder chil­dren, the slide is steeper, at $60^\circ$ to the ground, with its top at a height of 3 m. What should be the length of each slide?

In each case the slide is the hypotenuse and the height is the side oppo­site the given angle, so we use sine.

Slide for younger chil­dren:

$$\displaystyle \sin 30^\circ = \frac{1.5}{L_1} \implies L_1 = \frac{1.5}{1/2} = 3.$$

Slide for elder chil­dren:

$$\displaystyle \sin 60^\circ = \frac{3}{L_2} \implies L_2 = \frac{3}{\sqrt3/2} = \frac{6}{\sqrt3} = 2\sqrt3.$$

Check: $\displaystyle \frac{1.5}{3} = 0.5 = \sin 30^\circ$ and $\displaystyle \frac{3}{2\sqrt3} = \frac{\sqrt3}{2} = \sin 60^\circ$.

Answer: the slides should be 3 m and $2\sqrt3$ m $\approx 3.46$ m long.

Ques­tion 4#

The angle of ele­va­tion of the top of a tower from a point on the ground, 30 m away from the foot of the tower, is $30^\circ$. Find the height of the tower.

Solu­tion. Let the tower have height $h$ m. We know the adja­cent side (30 m) and want the oppo­site side, so we use tan­gent:

$$\displaystyle \tan 30^\circ = \frac{h}{30} \implies h = \frac{30}{\sqrt3} = \frac{30\sqrt3}{3} = 10\sqrt3.$$

Check: $\displaystyle \frac{10\sqrt3}{30} = \frac{\sqrt3}{3} = \frac{1}{\sqrt3} = \tan 30^\circ$.

Answer: the tower is $10\sqrt3$ m $\approx 17.32$ m high.

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, and its incli­na­tion with the ground is $60^\circ$. Find the length of the string, assum­ing there is no slack.

Solu­tion. The string $L$ is the hypotenuse and the height of 60 m is oppo­site the $60^\circ$ angle:

$$\displaystyle \sin 60^\circ = \frac{60}{L} \implies L = \frac{60}{\sqrt3/2} = \frac{120}{\sqrt3} = 40\sqrt3.$$

Check: $\displaystyle 40\sqrt3 \times \frac{\sqrt3}{2} = \frac{40 \times 3}{2} = 60$.

Answer: the string is $40\sqrt3$ m $\approx 69.28$ m long.

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.

Set­ting up. The angles are mea­sured from his eyes, which are 1.5 m above the ground. So the ver­ti­cal side of each tri­an­gle is

$$30 - 1.5 = 28.5 \text{ m}.$$

Let $d_1$ and $d_2$ be his hor­i­zon­tal dis­tances from the build­ing when the angle is $30^\circ$ and $60^\circ$.

Step 1.

$$\displaystyle \tan 30^\circ = \frac{28.5}{d_1} \implies d_1 = 28.5\sqrt3.$$

Step 2.

$$\displaystyle \tan 60^\circ = \frac{28.5}{d_2} \implies d_2 = \frac{28.5}{\sqrt3}.$$

Step 3. The dis­tance walked is the dif­fer­ence:

$$\displaystyle \begin{aligned} d_1 - d_2 &= 28.5\sqrt3 - \frac{28.5}{\sqrt3} = 28.5 \cdot \frac{3 - 1}{\sqrt3} \\ &= \frac{57}{\sqrt3} = 19\sqrt3. \end{aligned}$$

Check: $d_1 \approx 49.36$ m and $d_2 \approx 16.45$ m. Their dif­fer­ence is $32.91$ m, and $19 \times 1.732 \approx 32.91$.

Answer: he walked $19\sqrt3$ m $\approx 32.91$ m towards the build­ing.

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.

Set­ting up. Let the observer be $d$ m from the build­ing, and let the tower be $h$ m tall. The bot­tom of the tower is 20 m above the ground and its top is $(20 + h)$ m above the ground.

Step 1. Use the angle to the bot­tom of the tower:

$$\displaystyle \tan 45^\circ = \frac{20}{d} \implies d = 20.$$

Step 2. Use the angle to the top of the tower:

$$\displaystyle \tan 60^\circ = \frac{20 + h}{20} \implies 20 + h = 20\sqrt3 \implies h = 20(\sqrt3 - 1).$$

Check: $h \approx 20 \times 0.732 = 14.64$ m, and $\displaystyle \frac{20 + 14.64}{20} = 1.732$.

Answer: the tower is $20(\sqrt3 - 1)$ m $\approx 14.64$ m high.

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.

Set­ting up. Let the pedestal be $h$ m tall and the point be $d$ m from its foot.

Step 1.

$$\displaystyle \tan 45^\circ = \frac{h}{d} \implies d = h.$$

Step 2.

$$\displaystyle \tan 60^\circ = \frac{h + 1.6}{h} \implies \sqrt3\,h = h + 1.6 \implies h(\sqrt3 - 1) = 1.6.$$

Step 3. Ratio­nalise the denom­i­na­tor:

$$\displaystyle h = \frac{1.6}{\sqrt3 - 1} \times \frac{\sqrt3 + 1}{\sqrt3 + 1} = \frac{1.6(\sqrt3 + 1)}{2} = 0.8(\sqrt3 + 1).$$

Check: $h \approx 0.8 \times 2.732 \approx 2.19$ m, and $\displaystyle \frac{2.186 + 1.6}{2.186} \approx 1.732$.

Answer: the pedestal is $0.8(\sqrt3 + 1)$ m $\approx 2.19$ m high.

Ques­tion 9#

The angle of ele­va­tion of the top of a build­ing from the foot of a 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.

Set­ting up. Let the build­ing be $h$ m tall and the dis­tance between the feet of the two struc­tures be $d$ m.

Step 1. From the foot of the build­ing, look­ing at the 50 m tower:

$$\displaystyle \tan 60^\circ = \frac{50}{d} \implies d = \frac{50}{\sqrt3}.$$

Step 2. From the foot of the tower, look­ing at the build­ing:

$$\displaystyle \tan 30^\circ = \frac{h}{d} \implies h = \frac{50}{\sqrt3} \times \frac{1}{\sqrt3} = \frac{50}{3}.$$

Check: $d \approx 28.87$ m, and $\displaystyle \frac{16.67}{28.87} \approx 0.577 = \tan 30^\circ$.

Answer: the build­ing is $\displaystyle \frac{50}{3}$ m $\displaystyle = 16\tfrac23$ m $\approx 16.67$ m high.

Ques­tion 10#

Two poles of equal height stand oppo­site each other on either side of a road that is 80 m wide. From a point between them on the road, the angles of ele­va­tion of the tops of the poles are $60^\circ$ and $30^\circ$. Find the height of the poles and the dis­tances of the point from the poles.

Set­ting up. Let each pole be $h$ m tall. Let the point be $x$ m from the pole seen at $60^\circ$, so it is $(80 - x)$ m from the other pole.

Step 1.

$$\displaystyle \tan 60^\circ = \frac{h}{x} \implies h = \sqrt3\,x.$$

Step 2.

$$\displaystyle \tan 30^\circ = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt3}.$$

Step 3. Set the two expres­sions for $h$ equal:

$$\displaystyle \sqrt3\,x = \frac{80 - x}{\sqrt3} \implies 3x = 80 - x \implies x = 20.$$

So $h = 20\sqrt3$, and the other dis­tance is $80 - 20 = 60$ m.

Check: $\displaystyle \frac{20\sqrt3}{20} = \sqrt3 = \tan 60^\circ$ and $\displaystyle \frac{20\sqrt3}{60} = \frac{1}{\sqrt3} = \tan 30^\circ$.

Answer: each pole is $20\sqrt3$ m $\approx 34.64$ m high. The point is 20 m from the pole seen at $60^\circ$ and 60 m from the pole seen at $30^\circ$.

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 $30^\circ$. Find the height of the tower and the width of the canal.

Set­ting up. Let the tower be $h$ m tall and the canal $d$ m wide. The sec­ond point is far­ther from the tower, at a dis­tance of $(d + 20)$ m.

Step 1.

$$\displaystyle \tan 60^\circ = \frac{h}{d} \implies h = \sqrt3\,d.$$

Step 2.

$$\displaystyle \tan 30^\circ = \frac{h}{d + 20} \implies h = \frac{d + 20}{\sqrt3}.$$

Step 3. Set the two expres­sions equal:

$$\displaystyle \sqrt3\,d = \frac{d + 20}{\sqrt3} \implies 3d = d + 20 \implies d = 10, \quad h = 10\sqrt3.$$

Check: $\displaystyle \frac{10\sqrt3}{10} = \sqrt3$ and $\displaystyle \frac{10\sqrt3}{30} = \frac{1}{\sqrt3}$.

Answer: the tower is $10\sqrt3$ m $\approx 17.32$ m high, and the canal is 10 m wide.

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.

Set­ting up. Draw a hor­i­zon­tal line from the top of the build­ing to the tower. It meets the tower 7 m above the ground and splits the tower into a lower part of 7 m and an upper part of $(H - 7)$ m, where $H$ is the tow­er's height. Let $d$ be the dis­tance between the build­ing and the tower.

Step 1. The angle of depres­sion of the foot is $45^\circ$, so

$$\displaystyle \tan 45^\circ = \frac{7}{d} \implies d = 7.$$

Step 2. The angle of ele­va­tion of the top is $60^\circ$, so

$$\displaystyle \tan 60^\circ = \frac{H - 7}{7} \implies H - 7 = 7\sqrt3 \implies H = 7(1 + \sqrt3).$$

Check: $H \approx 7 \times 2.732 = 19.12$ m.

Answer: the cable tower is $7(\sqrt3 + 1)$ m $\approx 19.12$ m high.

Ques­tion 13#

As observed from the top of a light­house 75 m high above 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.

Set­ting up. The nearer ship is seen at the larger angle of depres­sion, $45^\circ$. Let the two ships be $d_1$ and $d_2$ m from the foot of the light­house. Because the angle of depres­sion equals the angle of ele­va­tion from the ship (alter­nate angles), we can work in the right tri­an­gles at sea level.

Step 1.

$$\displaystyle \tan 45^\circ = \frac{75}{d_1} \implies d_1 = 75.$$

Step 2.

$$\displaystyle \tan 30^\circ = \frac{75}{d_2} \implies d_2 = 75\sqrt3.$$

Step 3.

$$d_2 - d_1 = 75\sqrt3 - 75 = 75(\sqrt3 - 1).$$

Check: $75 \times 1.732 = 129.9$, and $129.9 - 75 = 54.9$.

Answer: the ships are $75(\sqrt3 - 1)$ m $\approx 54.9$ m apart.

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 her eyes at any instant is $60^\circ$. After some time, the angle of ele­va­tion reduces to $30^\circ$. Find the dis­tance trav­elled by the bal­loon dur­ing the inter­val.

Set­ting up. As in Ques­tion 6, mea­sure from eye level. The bal­loon is

$$88.2 - 1.2 = 87 \text{ m}$$

above her eyes. Let $d_1$ and $d_2$ be the hor­i­zon­tal dis­tances of the bal­loon at the two instants.

Step 1.

$$\displaystyle \tan 60^\circ = \frac{87}{d_1} \implies d_1 = \frac{87}{\sqrt3} = 29\sqrt3.$$

Step 2.

$$\displaystyle \tan 30^\circ = \frac{87}{d_2} \implies d_2 = 87\sqrt3.$$

Step 3. The angle decreased, so the bal­loon moved away from her. The dis­tance it trav­elled is

$$d_2 - d_1 = 87\sqrt3 - 29\sqrt3 = 58\sqrt3.$$

Check: $d_1 \approx 50.23$ m and $d_2 \approx 150.69$ m, so the dif­fer­ence is $100.46$ m, and $58 \times 1.732 \approx 100.46$.

Answer: the bal­loon trav­elled $58\sqrt3$ m $\approx 100.46$ 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 $30^\circ$, 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.

Set­ting up. Let the tower be $h$ m tall. Let the car be $d_1$ m from the foot at the first sight­ing and $d_2$ m at the sec­ond.

Step 1.

$$\displaystyle \tan 30^\circ = \frac{h}{d_1} \implies d_1 = \sqrt3\,h.$$

Step 2.

$$\displaystyle \tan 60^\circ = \frac{h}{d_2} \implies d_2 = \frac{h}{\sqrt3}.$$

Step 3. In 6 sec­onds the car cov­ers

$$\displaystyle d_1 - d_2 = \sqrt3\,h - \frac{h}{\sqrt3} = \frac{2h}{\sqrt3},$$

so its speed is

$$\displaystyle \frac{2h}{\sqrt3} \div 6 = \frac{h}{3\sqrt3} \text{ m/s}.$$

Step 4. The remain­ing dis­tance is $\displaystyle d_2 = \frac{h}{\sqrt3}$, so the time needed is

$$\displaystyle \frac{h}{\sqrt3} \div \frac{h}{3\sqrt3} = \frac{h}{\sqrt3} \times \frac{3\sqrt3}{h} = 3 \text{ s}.$$

Short­cut: $\displaystyle \frac{d_1}{d_2} = 3$, so the remain­ing dis­tance $d_2$ is half of the $2d_2$ the car has already cov­ered. At a uni­form speed, it there­fore takes half of 6 sec­onds.

Check: if $h = 10$ m, then $d_1 \approx 17.32$ m and $d_2 \approx 5.77$ m. The speed is $\displaystyle \frac{11.55}{6} \approx 1.925$ m/s, and $\displaystyle \frac{5.77}{1.925} \approx 3.0$ s.

Answer: the car takes 3 sec­onds to reach the foot of the tower.

In every ques­tion, the work comes down to find­ing the right tri­an­gle, choos­ing the ratio that links the known and unknown sides, and remem­ber­ing to sub­tract the observer's height when the angle is mea­sured from the eyes (Ques­tions 6 and 14).

Key terms#

Line of sight

The line from the observer's eye to the object being viewed.

Angle of ele­va­tion

The angle between the hor­i­zon­tal and the line of sight to an object above eye level.

Angle of depres­sion

The angle between the hor­i­zon­tal and the line of sight to an object below eye level.

Hor­i­zon­tal level

The line through the observer's eye par­al­lel to the ground; every angle in this exer­cise is mea­sured from it.

Hypotenuse

The side oppo­site the right angle, for exam­ple a rope, a kite string or a slide.

Tan­gent ratio

Oppo­site side divided by adja­cent side; the ratio used for most height and dis­tance prob­lems.

Ratio­nal­is­ing

Mul­ti­ply­ing by a con­ju­gate such as $\sqrt3 + 1$ to remove a surd from a denom­i­na­tor.

Com­mon ques­tions#

When should I use sine instead of tan­gent?#

Use sine when the hypotenuse (a rope, a string or a slide) is involved together with a height. Use tan­gent when the two sides you are work­ing with are a height and a hor­i­zon­tal dis­tance.

Why sub­tract the observer's height?#

The angle is mea­sured from the eyes, so the ver­ti­cal side of the tri­an­gle starts at eye level, not at the ground.

Is the angle of depres­sion inside the tri­an­gle?#

No, it lies between the hor­i­zon­tal line and the line of sight. It is equal to the angle of ele­va­tion at the other end, because they are alter­nate angles, so you can use that angle inside the tri­an­gle.

Should I leave answers with $\sqrt3$?#

Give the exact surd form first, as NCERT does. Add a dec­i­mal using $\sqrt3 \approx 1.732$ when the ques­tion asks for one.

Why does the height can­cel out in Ques­tion 15?#

Both dis­tances are mul­ti­ples of $h$, so their ratio depends only on the angles. The time there­fore does not depend on how tall the tower is.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.
3. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.