Some Applications of Trigonometry | Exercise 9.1

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

Angle of Elevation: The angle between the horizontal line of sight (from the eyes) and the line looking up to the top of an object. As the boy walks closer, the angle increases from 30° to 60°.

Key setup: The boy's eyes are at 1.5 m height, so the relevant vertical distance is the building height minus the eye level — this is the effective height used in the triangle.

We will use trigonometry to find the horizontal distances from the building at two different points and then find their difference.

Step 1 — Calculate effective height

The angle of elevation is measured from the boy's eyes, not from the ground. So we only need the height above his eye level.

First, let's find the height of the building from the boy's eye level. The building is 30 m tall. The boy is 1.5 m tall. So, we subtract the boy's height from the building's height.

$\text{Effective height} = 30 \text{ m} - 1.5 \text{ m}$

$= 28.5 \text{ m}$

$\boxed{28.5 \text{ m}}$

Diagram 1

Step 2 — Find initial distance

Let's call the initial distance CE as $x_1$ . Tangent Ratio: $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$ — used here because we know the height (opposite) and need the ground distance (adjacent).

We use the right-angled triangle AEC. The angle of elevation is $30^\circ$ . We know the opposite side AE is 28.5 m. We need to find the adjacent side CE.

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}$

$\tan(30^\circ) = \frac{AE}{CE}$

$\frac{1}{\sqrt{3}} = \frac{28.5}{x_1}$

$x_1 = 28.5 \times \sqrt{3}$

$\boxed{28.5\sqrt{3} \text{ m}}$

Step 3 — Find final distance

Now, let's call the final distance DE as $x_2$ . We use the right-angled triangle AED. The new angle of elevation is $60^\circ$ . The opposite side AE is still 28.5 m. We need to find the adjacent side DE.

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}$

$\tan(60^\circ) = \frac{AE}{DE}$

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

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

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

$x_2 = \frac{28.5 \times \sqrt{3}}{3}$

$x_2 = 9.5 \times \sqrt{3}$

$\boxed{9.5\sqrt{3} \text{ m}}$

Step 4 — Calculate distance walked

Since $x_1 > x_2$ (the boy is closer to the building at the second position), the distance walked = $x_1 - x_2$ .

The boy walked from his initial position to his final position. This distance is the difference between $x_1$ and $x_2$ . Let's calculate the distance walked.

$\text{Distance walked} = x_1 - x_2$

$= 28.5\sqrt{3} - 9.5\sqrt{3}$

$= (28.5 - 9.5)\sqrt{3}$

$= 19\sqrt{3}$

We can use $\sqrt{3} \approx 1.732$ .

$= 19 \times 1.732$

$= 32.908$

$\boxed{32.91 \text{ m (approx)}}$

Answer

The distance he walked towards the building is approximately 32.91 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 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.

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.