12. From the top of a high building, the angle of elevation of the top of a cable tower is and the angle of depression of its foot is . Determine the height of the tower.

Understand the Question Let be the building and be the cable tower standing on the same ground level . From the top of the building (point ), drawing a horizontal line creates two right-angled triangles: with angle of elevation from ground (alternate interior angle to angle of depression ). with angle of elevation . First, find the horizontal distance using . Then, find the upper height using . The total height is , where . Step 1 · Find the Horizontal Distance Let be the height of the building and be the cable tower. Draw . The angle of depression . In right Step 2 · Find the Upper Section of the Tower Since is a rectangle, and . In right Step 3 · Calculate the Total Height of the Tower The total height of the tower is Substituting Answer > Common Mistakes Angle of Depression Reference: Drawing the angle of depression with the vertical building line instead of the horizontal line of sight . Omitting the Base Height: Finding only and forgetting to add the lower segment to get the full tower height.

Some Applications of Trigonometry | Exercise 9.1

Question 12

From the top of a $7\text{ m}$ 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

Understand the Question

Let $\text{AB} = 7\text{ m}$ be the building and $\text{CD}$ be the cable tower standing on the same ground level $\text{AD}$ .
From the top of the building (point $\text{B}$ $B$ ), drawing a horizontal line $\text{BE} \parallel \text{AD}$ $BE ∥ AD$ creates two right-angled triangles:
- $\Delta \text{ABD}$ with angle of elevation from ground $\angle \text{BDA} = 45^\circ$ (alternate interior angle to angle of depression $\angle \text{EBD} = 45^\circ$ ).
- $\Delta \text{CBE}$ with angle of elevation $\angle \text{CBE} = 60^\circ$ .
First, find the horizontal distance $\text{AD} = \text{BE}$ using $\Delta \text{ABD}$ . Then, find the upper height $\text{CE}$ using $\Delta \text{CBE}$ . The total height is $\text{CD} = \text{CE} + \text{ED}$ , where $\text{ED} = \text{AB} = 7\text{ m}$ .

Step 1 · Find the Horizontal Distance

Let $\text{AB} = 7\text{ m}$ be the height of the building and $\text{CD}$ be the cable tower. Diagram 1

Draw $\text{BE} \parallel \text{AD}$ . The angle of depression $\angle \text{EBD} = 45^\circ \implies \angle \text{BDA} = 45^\circ$ .

In right $\Delta \text{ABD}$

\begin{aligned} \tan(\angle \text{BDA}) &= \dfrac{\text{AB}}{\text{AD}} \\[0.6em] \tan(45^\circ) &= \dfrac{7}{\text{AD}} \\[0.6em] 1 &= \dfrac{7}{\text{AD}} \\[0.6em] \text{AD} &= 7\text{ m} \end{aligned}

Step 2 · Find the Upper Section of the Tower

Since $\text{ABED}$ is a rectangle, $\text{BE} = \text{AD} = 7\text{ m}$ and $\text{ED} = \text{AB} = 7\text{ m}$ .

In right $\Delta \text{CBE}$

\begin{aligned} \tan(\angle \text{CBE}) &= \dfrac{\text{CE}}{\text{BE}} \\[0.6em] \tan(60^\circ) &= \dfrac{\text{CE}}{7} \\[0.6em] \sqrt{3} &= \dfrac{\text{CE}}{7} \\[0.6em] \text{CE} &= 7\sqrt{3}\text{ m} \end{aligned}

Step 3 · Calculate the Total Height of the Tower

The total height of the tower $\text{CD}$ is $\text{CE} + \text{ED}$

\begin{aligned} \text{CD} &= \text{CE} + \text{ED} \\ &= 7\sqrt{3} + 7 \\ &= 7(1 + \sqrt{3})\text{ m} \end{aligned}

Substituting $\sqrt{3} \approx 1.732$

\begin{aligned} \text{CD} &= 7(1 + 1.732) \\ &= 7(2.732) \\ &= 19.124\text{ m} \end{aligned}

Answer

$7(1 + \sqrt{3})\text{ m} \approx 19.124\text{ m}$

Common Mistakes

Angle of Depression Reference: Drawing the $45^\circ$ angle of depression with the vertical building line $\text{AB}$ instead of the horizontal line of sight $\text{BE}$ .
Omitting the Base Height: Finding only $\text{CE} = 7\sqrt{3}\text{ m}$ and forgetting to add the lower segment $\text{ED} = 7\text{ m}$ to get the full tower height.

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 3 m, 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 $60\text{ m}$ 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.

A $1.5\text{ m}$ tall boy is standing at some distance from a $30\text{ m}$ 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, $1.6\text{ m}$ 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 $7\text{ m}$ 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 $75\text{ m}$ 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 $1.2\text{ m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2\text{ m}$ 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.