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A kite is flying at a height of $60$ m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $30^\circ$. From the bottom of the same building, the angle of elevation of kite is $45^\circ$. Find the length of the string and height of roof from the ground. (Use $\sqrt{3} = 1.73$)
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Let K be the position of kite and TR is the height of building.
Correct figure (I) (1 Mark)
$\therefore \tan 45^\circ = 1 = \frac{60}{GT}$ (II) (1 Mark)
GT = $60$ m
Also, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{KS}{SR}$ (III) (1 Mark)
$\Rightarrow KS = 20\sqrt{3}$ m or $34.6$ m (IV) (1/2 Mark)
Hence, TR = $(60 - 20\sqrt{3})$ m = $60 - 34.6 = 25.4$ m (V) (1/2 Mark)
Also, $\sin 30^\circ = \frac{1}{2} = \frac{KS}{KR} = \frac{20\sqrt{3}}{KR}$ (VI) (1/2 Mark)
$\Rightarrow KR = 40\sqrt{3} = 69.2$ m (VII) (1/2 Mark)
$\therefore$ The length of the string = $69.2$ m and height of roof from the ground = $25.4$ m
Correct figure (I) (1 Mark)
$\therefore \tan 45^\circ = 1 = \frac{60}{GT}$ (II) (1 Mark)
GT = $60$ m
Also, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{KS}{SR}$ (III) (1 Mark)
$\Rightarrow KS = 20\sqrt{3}$ m or $34.6$ m (IV) (1/2 Mark)
Hence, TR = $(60 - 20\sqrt{3})$ m = $60 - 34.6 = 25.4$ m (V) (1/2 Mark)
Also, $\sin 30^\circ = \frac{1}{2} = \frac{KS}{KR} = \frac{20\sqrt{3}}{KR}$ (VI) (1/2 Mark)
$\Rightarrow KR = 40\sqrt{3} = 69.2$ m (VII) (1/2 Mark)
$\therefore$ The length of the string = $69.2$ m and height of roof from the ground = $25.4$ m