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…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 5 Marks · March 2026 · Standard
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495 Marks · March 2026 · Standard
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) $\Rightarrow GT = 60$ m (III) (1 Mark) Also, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{KS}{SR}$ (IV) ($\frac{1}{2}$ Mark) $\Rightarrow KS = 20\sqrt{3}$ m or $34.6$ m (V) ($\frac{1}{2}$ Mark) Hence, $TR = (60 - 20\sqrt{3})$ m $= 60 - 34.6 = 25.4$ m (VI) ($\frac{1}{2}$ Mark) Also, $\sin 30^\circ = \frac{1}{2} = \frac{KS}{KR}$ (VII) ($\frac{1}{2}$ Mark) $\Rightarrow KR = 40\sqrt{3} = 69.2$ m $\therefore$ The length of the string $= 69.2$ m and height of roof from the ground $= 25.4$ m
