Exams › SSC CGL (Prelims) › General

A vertical tower of height \(H\) stands on the ground. From point A on the ground, the angle of elevation to the top is \(45^\circ\). From point B, located 15 meters away from A along the line extending from the base, the angle of elevation is \(30^\circ\). What is the height of the tower?

1. 15(\sqrt{3} - 1) m
2. 15(\sqrt{3} + 1)/2 m
3. 20(\sqrt{3} - 1) m
4. 20(\sqrt{3} + 1) m

Correct answer: 15(\sqrt{3} + 1)/2 m

Solution

Let the distance from A to the tower’s base be \(x\). From A, \(\tan 45^\circ=H/x\), so \(H=x\). From B, \(\tan 30^\circ=H/(x+15)=1/\sqrt{3}\). Substituting \(H=x\) gives \(x/(x+15)=1/\sqrt{3}\), which yields \(H=\frac{15(\sqrt{3}+1)}{2}\) m.

Related SSC CGL (Prelims) General questions

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