NCERT Solutions For Class 10 Maths: Some Application of Trigonometry

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 level is $30^\circ$ (see Fig. 9.11).

Solution-
1.Let the height of the vertical pole be $h$. The length of the rope (the line of sight or hypotenuse) is 20 m. The angle of elevation is $30^\circ$.

2.In the right triangle formed, we use the $\sin$ trigonometric ratio as it relates the opposite side (height) and the hypotenuse (rope length).
$\sin 30^\circ = \frac{\text{Height of the pole}}{\text{Length of the rope}}$

3.$\frac{1}{2} = \frac{h}{20}$

4.$h = \frac{20}{2} = 10$ m.

5.The height of the pole is 10 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.

Solution-
1.Let $h$ be the height of the standing part of the tree (Opposite side) and $L$ be the length of the broken part (Hypotenuse). The distance from the foot of the tree (Adjacent side) is 8 m, and the angle of elevation is $30^\circ$. The total height of the tree is $h + L$.

2.Find the standing part $h$ using $\tan 30^\circ$:
$\tan 30^\circ = \frac{h}{8}$

3.$h = 8 \times \frac{1}{\sqrt{3}} = \frac{8}{\sqrt{3}}$ m.

4.Find the broken part $L$ using $\cos 30^\circ$:
$\cos 30^\circ = \frac{8}{L}$

5.$\frac{\sqrt{3}}{2} = \frac{8}{L}$

6.$L = \frac{16}{\sqrt{3}}$ m.

7.Total height of the tree $= h + L = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}}$

8.Rationalizing the denominator: $\frac{24\sqrt{3}}{3} = 8\sqrt{3}$ 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 3m, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?

Solution-
1.Case 1: Younger children. Height $H_1 = 1.5$ m, angle of inclination $30^\circ$. Let the slide length be $L_1$.
$\sin 30^\circ = \frac{1.5}{L_1}$

2.$\frac{1}{2} = \frac{1.5}{L_1}$

3.$L_1 = 3$ m.

4.Case 2: Elder children. Height $H_2 = 3$ m, angle of inclination $60^\circ$. Let the slide length be $L_2$.
$\sin 60^\circ = \frac{3}{L_2}$

5.$\frac{\sqrt{3}}{2} = \frac{3}{L_2}$

6.$L_2 = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$ m.

7.The length of the slide for younger children is 3 m, and for elder children is $2\sqrt{3}$ m.

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.

Solution-
1.Let $h$ be the height of the tower (Opposite side). The distance from the foot of the tower (Adjacent side) is 30 m. The angle of elevation is $30^\circ$.

2.We use the $\tan$ trigonometric ratio:
$\tan 30^\circ = \frac{h}{30}$

3.$\frac{1}{\sqrt{3}} = \frac{h}{30}$

4.$h = \frac{30}{\sqrt{3}}$

5.Rationalizing: $h = \frac{30\sqrt{3}}{3} = 10\sqrt{3}$ 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.

Solution-
1.The height of the kite (Opposite side) is 60 m. The inclination of the string (angle of elevation) is $60^\circ$. Let $L$ be the length of the string (Hypotenuse).

2.We use the $\sin$ trigonometric ratio:
$\sin 60^\circ = \frac{60}{L}$

3.$\frac{\sqrt{3}}{2} = \frac{60}{L}$

4.$L = \frac{120}{\sqrt{3}}$

5.Rationalizing: $L = \frac{120\sqrt{3}}{3} = 40\sqrt{3}$ m.

6.The length of the string is $40\sqrt{3}$ 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.

Solution-
1.The effective height of the building above the observer's eye level (AE) is $30 \text{ m} - 1.5 \text{ m} = 28.5$ m. Let $x_1$ be the initial distance and $x_2$ be the final distance from the building. The distance walked is $d = x_1 - x_2$.

2.In the right triangle with the final angle of elevation $60^\circ$:
$\tan 60^\circ = \frac{28.5}{x_2}$

3.$x_2 = \frac{28.5}{\sqrt{3}}$ m.

4.In the right triangle with the initial angle of elevation $30^\circ$:
$\tan 30^\circ = \frac{28.5}{x_1}$

5.$x_1 = 28.5\sqrt{3}$ m.

6.Distance walked $d = x_1 - x_2 = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = 28.5 \left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{57}{\sqrt{3}}$.

7.Rationalizing: $d = \frac{57\sqrt{3}}{3} = 19\sqrt{3}$ 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.

Solution-
1.Let the height of the building be $20$ m and the height of the transmission tower be $h$. Let $x$ be the distance from the point on the ground to the foot of the building.

2.For the angle of elevation of the bottom of the tower ($45^\circ$):
$\tan 45^\circ = \frac{20}{x}$

3.$1 = \frac{20}{x}$, so $x = 20$ m.

4.For the angle of elevation of the top of the tower ($60^\circ$). The total height is $20 + h$.
$\tan 60^\circ = \frac{20 + h}{x}$

5.$\sqrt{3} = \frac{20 + h}{20}$

6.$20\sqrt{3} = 20 + h$

7.$h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1)$ m.

8.The height of the tower is $20(\sqrt{3} - 1)$ 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.

Solution-
1.Let the height of the pedestal be $h$ and the distance from the point on the ground be $x$. The statue height is 1.6 m.

2.For the angle of elevation of the top of the pedestal ($45^\circ$):
$\tan 45^\circ = \frac{h}{x}$

3.$1 = \frac{h}{x}$, so $x = h$.

4.For the angle of elevation of the top of the statue ($60^\circ$). The total height is $h + 1.6$.
$\tan 60^\circ = \frac{h + 1.6}{x}$

5.Substituting $x=h$:
$\sqrt{3} = \frac{h + 1.6}{h}$

6.$h\sqrt{3} = h + 1.6$

7.$h(\sqrt{3} - 1) = 1.6$

8.$h = \frac{1.6}{\sqrt{3} - 1}$

9.Rationalizing: $h = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{1.6(\sqrt{3} + 1)}{3 - 1} = 0.8(\sqrt{3} + 1)$ 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.

Solution-
1.Let $H_T$ be the tower height (50 m) and $H_B$ be the building height ($h$). Let $x$ be the distance between the two structures.

2.In the right triangle involving the tower (angle $60^\circ$):
$\tan 60^\circ = \frac{50}{x}$

3.$\sqrt{3} = \frac{50}{x}$, so $x = \frac{50}{\sqrt{3}}$ m.

4.In the right triangle involving the building (angle $30^\circ$):
$\tan 30^\circ = \frac{h}{x}$

5.$\frac{1}{\sqrt{3}} = \frac{h}{50/\sqrt{3}}$

6.$h = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3}$ m.

7.The height of the building is $50/3$ m.

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.

Solution-
1.Let $h$ be the height of the poles. The width of the road is 80 m. Let $x$ be the distance of the observation point from the first pole, so the distance from the second pole is $80 - x$.

2.In the triangle corresponding to the $60^\circ$ angle of elevation:
$\tan 60^\circ = \frac{h}{x}$

3.$h = x\sqrt{3}$ (1)

4.In the triangle corresponding to the $30^\circ$ angle of elevation:
$\tan 30^\circ = \frac{h}{80 - x}$

5.$\frac{1}{\sqrt{3}} = \frac{h}{80 - x}$

6.Substitute $h$ from (1) into (5):
$\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{80 - x}$

7.$80 - x = 3x$

8.$4x = 80$, so $x = 20$ m.

9.The distances of the point from the poles are 20 m and $80 - 20 = 60$ m.

10.Calculate height $h$: $h = x\sqrt{3} = 20\sqrt{3}$ m.

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 joing 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.

Solution-
1.Let $h$ be the height of the TV tower and $x$ be the width of the canal. The total distance for the $30^\circ$ angle of elevation is $x + 20$ m.

2.In the triangle corresponding to the $60^\circ$ angle:
$\tan 60^\circ = \frac{h}{x}$

3.$h = x\sqrt{3}$ (1)

4.In the triangle corresponding to the $30^\circ$ angle:
$\tan 30^\circ = \frac{h}{x + 20}$

5.$\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$, so $h\sqrt{3} = x + 20$ (2)

6.Substitute $h$ from (1) into (2):
$(x\sqrt{3})\sqrt{3} = x + 20$

7.$3x = x + 20$

8.$2x = 20$, so $x = 10$ m.

9.The width of the canal is 10 m.

10.Calculate height $h$: $h = x\sqrt{3} = 10\sqrt{3}$ 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.

Solution-
1.Let the height of the building be 7 m. Let $x$ be the distance between the building and the cable tower.

2.The angle of depression of the tower's foot is $45^\circ$, which means the angle of elevation from the foot of the building is also $45^\circ$.
$\tan 45^\circ = \frac{7}{x}$

3.$1 = \frac{7}{x}$, so $x = 7$ m.

4.The angle of elevation of the top of the tower from the top of the building is $60^\circ$. Let $h$ be the height of the tower above the building.
$\tan 60^\circ = \frac{h}{x}$

5.$\sqrt{3} = \frac{h}{7}$

6.$h = 7\sqrt{3}$ m.

7.The height of the tower $= 7 + h = 7 + 7\sqrt{3} = 7(1 + \sqrt{3})$ 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.

Solution-
1.The height of the lighthouse is 75 m. Let $x$ be the distance to the nearer ship and $y$ be the distance between the ships.

2.For the nearer ship, the angle of depression is $45^\circ$, so the angle of elevation is $45^\circ$.
$\tan 45^\circ = \frac{75}{x}$

3.$x = 75$ m.

4.For the farther ship, the angle of depression is $30^\circ$, so the angle of elevation is $30^\circ$. The total distance is $x + y$.
$\tan 30^\circ = \frac{75}{x + y}$

5.$\frac{1}{\sqrt{3}} = \frac{75}{75 + y}$

6.$75 + y = 75\sqrt{3}$

7.$y = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)$ m.

8.The distance between the two ships is $75(\sqrt{3} - 1)$ 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.

Solution-
1.The effective height of the balloon above the girl's eye level is $H = 88.2 \text{ m} - 1.2 \text{ m} = 87$ m. Let $x_1$ and $x_2$ be the horizontal distances corresponding to $60^\circ$ and $30^\circ$, respectively. The distance travelled is $d = x_2 - x_1$.

2.For the initial angle of elevation ($60^\circ$):
$\tan 60^\circ = \frac{87}{x_1}$

3.$x_1 = \frac{87}{\sqrt{3}}$ m.

4.For the final angle of elevation ($30^\circ$):
$\tan 30^\circ = \frac{87}{x_2}$

5.$x_2 = 87\sqrt{3}$ m.

6.Distance travelled $d = 87\sqrt{3} - \frac{87}{\sqrt{3}} = 87 \left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{174}{\sqrt{3}}$.

7.Rationalizing: $d = \frac{174\sqrt{3}}{3} = 58\sqrt{3}$ 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.

Solution-
1.Let $h$ be the height of the tower. The initial angle of depression is $30^\circ$ and the final angle is $60^\circ$. By alternate angles, the angles of elevation from the car positions are $30^\circ$ and $60^\circ$.

2.Let $x$ be the remaining distance to the foot of the tower (at $60^\circ$), and $y$ be the distance covered in 6 seconds.

3.In the $60^\circ$ triangle:
$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3}$.

4.In the $30^\circ$ triangle:
$\tan 30^\circ = \frac{h}{x + y}$

5.$\frac{1}{\sqrt{3}} = \frac{h}{x + y} \implies h\sqrt{3} = x + y$.

6.Substitute $h$: $(x\sqrt{3})\sqrt{3} = x + y \implies 3x = x + y$.

7.$y = 2x$.

8.Since the car is moving at uniform speed, distance is proportional to time. Since the distance $y$ is covered in 6 seconds, and $y$ is twice the distance $x$, the time required to cover distance $x$ must be half the time required to cover distance $y$.

9.Time taken for distance $x = \frac{6 \text{ seconds}}{2} = 3$ seconds.