24
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. Tower is supported by wires from a point O.
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 $30^{\circ}$ and the angle of elevation of the top of Section A is $45^{\circ}$.
Based on the above information, answer the following questions :
(i) Find the length of the wire from the point O to the top of Section B.
(ii) Find the distance AB.
OR
Find the area of $\triangle OPB$.
(iii) Find the height of the Section A from the base of the tower.
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 $30^{\circ}$ and the angle of elevation of the top of Section A is $45^{\circ}$.
Based on the above information, answer the following questions :
(i) Find the length of the wire from the point O to the top of Section B.
(ii) Find the distance AB.
OR
Find the area of $\triangle OPB$.
(iii) Find the height of the Section A from the base of the tower.
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(i) In $\triangle OBP$, $\cos 30^{\circ} = \frac{OP}{OB}$
$\frac{\sqrt{3}}{2} = \frac{36}{OB} \Rightarrow OB = \frac{72}{\sqrt{3}} = 24\sqrt{3}$ cm
(ii)In $\triangle OBP$, $\tan 30^{\circ} = \frac{PB}{36} \Rightarrow PB = \frac{36}{\sqrt{3}} = 12\sqrt{3}$
In $\triangle OAP$, $\tan 45^{\circ} = \frac{AP}{36} \Rightarrow AP = 36$ cm
$AB = AP - PB = 36 - 12\sqrt{3} = 12(3 - \sqrt{3})$ cm
OR
(ii)Area of $\triangle OPB = \frac{1}{2} \times OP \times PB$
$= \frac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}$ cm$^2$
(iii) $AP = 36$ cm
$\frac{\sqrt{3}}{2} = \frac{36}{OB} \Rightarrow OB = \frac{72}{\sqrt{3}} = 24\sqrt{3}$ cm
(ii)In $\triangle OBP$, $\tan 30^{\circ} = \frac{PB}{36} \Rightarrow PB = \frac{36}{\sqrt{3}} = 12\sqrt{3}$
In $\triangle OAP$, $\tan 45^{\circ} = \frac{AP}{36} \Rightarrow AP = 36$ cm
$AB = AP - PB = 36 - 12\sqrt{3} = 12(3 - \sqrt{3})$ cm
OR
(ii)Area of $\triangle OPB = \frac{1}{2} \times OP \times PB$
$= \frac{1}{2} \times 36 \times 12\sqrt{3} = 216\sqrt{3}$ cm$^2$
(iii) $AP = 36$ cm