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Some Applications Of Trigonometry

Question

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The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30^o. The distance of the car from the base of the tower (in m.) is

$25 \sqrt{3}$

$50 \sqrt{3}$

$75 \sqrt{3}$

150

Solution

$75 \sqrt{3}$

Let AB be the tower of height 75 m and C be position of the car.

$∆ ABC, \cot 30 ° = \frac{AC}{AB} \Rightarrow AC = AB \cot 30 ° \Rightarrow AC = 75 m x \sqrt{3} \Rightarrow AC = 75 \sqrt{3} m Thus, the distance of the car from the base of the tower is 75 \sqrt{3} m .$

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Some Applications Of Trigonometry

The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30^o. The distance of the car from the base of the tower (in m.) is

$25 \sqrt{3}$

$50 \sqrt{3}$

$75 \sqrt{3}$

150

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For Daily Free Study Material Join wiredfaculty

Some Applications Of Trigonometry

The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30o. The distance of the car from the base of the tower (in m.) is 253 503 753 150

Some More Questions From Some Applications of Trigonometry Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30^o. The distance of the car from the base of the tower (in m.) is

$25 \sqrt{3}$

$50 \sqrt{3}$

$75 \sqrt{3}$

150