Question

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections A and B. The tower is supported by wires from a point O. The distance between the base of the tower and point O is 36 cm. From point O, the angle of elevation of the top of the section B is 30° and the angle of elevation of the top of section A is 45°.

Based on the given information, answer the following questions.

1. Find the length of the wire from the point O to the top of section B.
2. Find the distance AB.
3. Find the area of ΔOPB.
4. Find the height of section A from the base of the tower.

Answer 1

i. Length of wire from O to top of section B

Let P be the base of the tower. Then OP = 36 cm.

In right \(\triangle OPB\), with \(\angle POB = 30^\circ\):

\[\cos 30^\circ = \frac{OP}{OB}\]

\[\Rightarrow \frac{\sqrt{3}}{2} = \frac{36}{OB}\]

\[\Rightarrow OB = \frac{36 \times 2}{\sqrt{3}}\]

\[= 24\sqrt{3} \text{ cm}\]

ii. Distance AB

Let A be the top of section A and B the top of section B. Then AP is the total height of the tower and BP is the height of section B.

In right \(\triangle OPA\), with \(\angle POA = 45^\circ\):

\[\tan 45^\circ = \frac{AP}{OP}\]

\[\Rightarrow 1 = \frac{AP}{36}\]

\[\Rightarrow AP = 36 \text{ cm}\]

In right \(\triangle OPB\), with \(\angle POB = 30^\circ\):

\[\tan 30^\circ = \frac{BP}{OP}\]

\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{BP}{36}\]

\[\Rightarrow BP = \frac{36}{\sqrt{3}}\]

\[= 12\sqrt{3} \text{ cm}\]

Therefore, \[AB = AP - BP\]

\[= 36 - 12\sqrt{3}\]

\[= 12\sqrt{3}(\sqrt{3} - 1) \text{ cm}\]

iii. Area of \(\triangle OPB\)

\[\text{Area} = \frac{1}{2} \times OP \times BP\]

\[= \frac{1}{2} \times 36 \times 12\sqrt{3}\]

\[= 216\sqrt{3} \text{ cm}^2\]

iv. Height of section A from the base of the tower

From (ii), AP = 36 cm. Thus, the height of section A from the base P of the tower is \[AP = 36 \text{ cm}\].