Some Applications of Trigonometry | Exercise 9.1

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

We can use trigonometry to find the unknown length in a right-angled triangle.

Step 1 — Understand the situation

Let's draw a picture. The kite is 60 m high. This is the perpendicular height. Angle of Inclination: The angle a line (here, the string) makes with the horizontal ground. It is measured upward from the ground.

The string makes an angle of $60^\circ$ with the ground. We need to find the length of the string. This forms a right-angled triangle.

Diagram 1

Step 2 — Apply trigonometry

Let the height of the kite be $h$ . So, $h = \text{AB} = \mathbf{60 \text{ m}}$ . Let the length of the string be $L$ . So, $L = \text{AC}$ . The angle of inclination is $\theta = \mathbf{60^\circ}$ . In the right triangle ABC, AB is opposite to angle $\theta$ . AC is the hypotenuse. Sine Ratio: $\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$ — the height (opposite) and string length (hypotenuse) are connected through the angle.

We use the sine ratio. $\sin(\theta) = \text{Opposite} / \text{Hypotenuse}$

$\sin(60^\circ) = \frac{60}{L}$

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$ .

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

Now, we solve for $L$ .

$L \times \sqrt{3} = 60 \times 2$

$L \times \sqrt{3} = 120$

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

Rationalizing: Multiply numerator and denominator by $\sqrt{3}$ to clear the surd from the denominator.

To simplify, we multiply the numerator and denominator by $\sqrt{3}$ .

$L = \frac{120 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

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

$L = 40 \sqrt{3}$

$\boxed{40\sqrt{3} \text{ m}}$

Answer

(i) The length of the string is $40\sqrt{3} \text{ m}$ .

More questions in 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 $30^\circ$ (see Fig. 9.11).

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.

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?

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.

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.

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.

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 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.

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

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 $60^\circ$ . 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.

Q12

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.

Q13

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $30^\circ$ 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.

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 $60^\circ$ . 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.

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 $30^\circ$ , 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.