Question 69

A pilot in an aeroplane at an altitude of 200 m observes two points lying on either side of a river. If the angles of depression of the two points be 45° and 60°, then the width of the river is

Solution

Given : A is the aeroplane and AD = 200 m

To find : Width of river = BC = ?

Solution : In $$\triangle$$ ADC

=> $$tan(60^\circ)=\frac{AD}{DC}$$

=> $$\sqrt{3}=\frac{200}{DC}$$

=> $$DC=\frac{200}{\sqrt{3}}$$ m

Similarly, in $$\triangle$$ ABD

=> $$tan(45^\circ)=\frac{AD}{DB}$$

=> $$1=\frac{200}{DB}$$

=> $$DB=200$$ m

$$\therefore$$ BC = BD + DC

= $$(200 + \frac{200}{\sqrt{3}})$$ m

=> Ans - (A)

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