Maths · Trigonometrical identities and trigonometrical functions
A man from top of a 100 meters high towers sees a car moving towards the tower a
A man from top of a 100 meters high towers sees a car moving towards the tower at an angle of depression of \( 30^{0} \) After some time, the angle of depression becomes \( 60^{0} . \) The distance (in meters) travelled by the car during this time is
- A. \( 100 \sqrt{3} \) 3
- B. \( \frac{200 \sqrt{3}}{3} \)
- C. \( \frac{100 \sqrt{3}}{3} \)
- D. \( 200 \sqrt{3} \)
Step-by-step solution
Let the height of the tower be 100 m. Initially, angle of depression is 30°, so distance from tower base to car is d1 = 100 / tan(30°) = 100√3 m. Later, angle is 60°, so d2 = 100 / tan(60°) = 100/√3 = 100√3/3 m. Distance travelled = d1 - d2 = 100√3 - 100√3/3 = 200√3/3 m.
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