Tejas is standing at the top of a building and observes a car at an angle of depression of 30° as it approaches the…
CBSE Class 10 Maths PYQ · Applications of Trig · Double Triangle · 4 Marks · March 2026 · Standard
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344 Marks · March 2026 · Standard
Tejas is standing at the top of a building and observes a car at an angle of depression of $30^\circ$ as it approaches the base of the building at a uniform speed. 6 seconds later, the angle of depression increases to $60^\circ$, and at that moment, the car is 25 m away from the building. Based on the information given above, answer the following questions : (i) What is the height of the building? (ii) What is the distance between the two positions of the car ? (iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point ? OR (iii) (b) What is the distance of the observer from the car when it makes an angle of $60^\circ$?
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(i) In $\triangle ABC$, $\tan 60^\circ = \sqrt{3} = \frac{AB}{25}$ (1/2 Mark) $\Rightarrow AB = 25\sqrt{3}$ (1/2 Mark) $\therefore$ Height of building = $25\sqrt{3}$ m (1 Mark) (ii) In $\triangle ABD$, $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD}$ (1/2 Mark) $\Rightarrow BD = 75$ (1/2 Mark) $\therefore$ Distance between two positions of car = $75 - 25 = 50$ m (1 Mark) (iii) (a) Time taken to cover the distance of 50 m = 6 sec (1 Mark) Time taken to cover the distance of 75 m = $\frac{6}{50} \times 75$ $= 9 \text{ sec}$ (1 Mark) OR (iii) (b) In $\triangle ABC$, $\cos 60^\circ = \frac{BC}{AC}$ (1 Mark) $\Rightarrow \frac{1}{2} = \frac{25}{AC}$ AC = 50 (1 Mark) $\therefore$ Distance of the observer from car when it makes the angle of $60^\circ = 50 \text{ m}$
