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प्रश्न
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.
- Find the length of the wire from the point O to the top of section B.
- Find the distance AB.
- Find the area of ΔOPB.
- Find the height of section A from the base of the tower.
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उत्तर
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 \[AP = 36 \text{ cm}\].
