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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' (as shown in figure).
Distance between the base of the tower and point 'O' is $6$ m. 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 $60^\circ$.
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 length of the wire from the point 'O' to the top of section 'A'.
(iii) (a) Find the distance AB.
OR
(iii) (b) Find the area of $\triangle OPB$.
Distance between the base of the tower and point 'O' is $6$ m. 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 $60^\circ$.
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 length of the wire from the point 'O' to the top of section 'A'.
(iii) (a) Find the distance AB.
OR
(iii) (b) Find the area of $\triangle OPB$.
Show SolutionHide Solution↓
(i) $\cos 30^\circ = \frac{6}{OB} = \frac{\sqrt{3}}{2}$ (I) (1/2 Mark)
$\Rightarrow OB = \frac{12}{\sqrt{3}}$ or $4\sqrt{3}$ m (II) (1/2 Mark)
(ii) $\cos 60^\circ = \frac{6}{OA} = \frac{1}{2}$ (I) (1/2 Mark)
$\Rightarrow OA = 12$ m (II) (1/2 Mark)
(iii) (a) $\tan 30^\circ = \frac{BP}{6} = \frac{1}{\sqrt{3}}$ (I) (1 Mark)
$\Rightarrow BP = 2\sqrt{3}$ m
$\tan 60^\circ = \frac{AP}{6} = \sqrt{3}$
$\Rightarrow AP = 6\sqrt{3}$ m (II) (1/2 Mark)
$AB = AP – BP = 6\sqrt{3} – 2\sqrt{3} = 4\sqrt{3}$ m (III) (1/2 Mark)
OR
(iii) (b) $\tan 30^\circ = \frac{BP}{6} = \frac{1}{\sqrt{3}}$ (I) (1 Mark)
$\Rightarrow BP = 2\sqrt{3}$ m
$ar(\triangle OPB) = \frac{1}{2} \times BP \times OP$
$= \frac{1}{2} \times 2\sqrt{3} \times 6 = 6\sqrt{3}$ m$^2$ (II) (1 Mark)
$\Rightarrow OB = \frac{12}{\sqrt{3}}$ or $4\sqrt{3}$ m (II) (1/2 Mark)
(ii) $\cos 60^\circ = \frac{6}{OA} = \frac{1}{2}$ (I) (1/2 Mark)
$\Rightarrow OA = 12$ m (II) (1/2 Mark)
(iii) (a) $\tan 30^\circ = \frac{BP}{6} = \frac{1}{\sqrt{3}}$ (I) (1 Mark)
$\Rightarrow BP = 2\sqrt{3}$ m
$\tan 60^\circ = \frac{AP}{6} = \sqrt{3}$
$\Rightarrow AP = 6\sqrt{3}$ m (II) (1/2 Mark)
$AB = AP – BP = 6\sqrt{3} – 2\sqrt{3} = 4\sqrt{3}$ m (III) (1/2 Mark)
OR
(iii) (b) $\tan 30^\circ = \frac{BP}{6} = \frac{1}{\sqrt{3}}$ (I) (1 Mark)
$\Rightarrow BP = 2\sqrt{3}$ m
$ar(\triangle OPB) = \frac{1}{2} \times BP \times OP$
$= \frac{1}{2} \times 2\sqrt{3} \times 6 = 6\sqrt{3}$ m$^2$ (II) (1 Mark)