SECTION E This section has 3 case study based questions carrying 4 marks each. Case Study -1 The International Kite…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 4 Marks · July 2025 · Standard
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284 Marks · July 2025 · Standard
SECTION E This section has $3$ case study based questions carrying $4$ marks each. Case Study -1 The International Kite Festival takes place every year on $14^{th}$ January. The main attractions of the festival include national and international Kite Flyers' Parade, kite flying, traditional stalls etc. On this day, few kite flyers, had assembled at a point 'O' on the ground. The position of $3$ kites A, B, C was such that A and B were at the same vertical height of $40$ m from the ground level. The angles of elevation of A, B and C from O were $60^{\circ}, 45^{\circ}$ and $30^{\circ}$ respectively. A vertical tower, SD has been erected at point S and a camera is set at the top of the tower for photography. Based on the information given above, answer the following questions : (i) What is the length of the string of the kite at A? (ii) If the length of the string of kite at C is $40$ m, then find the height of that kite C from the ground. (iii) (a) What is the horizontal distance between the kites at A and B? OR (iii) (b) If the angle of depression of the kite at A is $30^{\circ}$ from the camera at D and the distance between A and D is $60$ m, then find the height of the tower.
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(i) $\sin 60^{\circ} = \frac{40}{OA} = \frac{\sqrt{3}}{2}$ $\Rightarrow OA = \frac{80}{\sqrt{3}}$ m or $\frac{80\sqrt{3}}{3}$ m (ii) $\sin 30^{\circ} = \frac{RC}{40} = \frac{1}{2}$ $\Rightarrow RC = 20$ m (iii) (a) $\tan 45^{\circ} = \frac{40}{OQ} = 1$ $\Rightarrow OQ = 40$ m Also, $\tan 60^{\circ} = \frac{40}{OP} = \sqrt{3}$ $\Rightarrow OP = \frac{40}{\sqrt{3}}$ m or $\frac{40\sqrt{3}}{3}$ m AB = PQ = $\left(40 + \frac{40}{\sqrt{3}}\right)$ m or $\left(40 + \frac{40\sqrt{3}}{3}\right)$ m OR (b) $\sin 30^{\circ} = \frac{h}{60} = \frac{1}{2}$ $\Rightarrow h = 30$ m Height of the tower = $40 + 30 = 70$ m
