Question 141

Two points A and B are on the ground and on opposite sides of a tower. A is closer to the foot of tower by 42 m than B. If the angles of elevation ofthe top of the tower, as observed from A and B are $$60^\circ$$ and $$45^\circ$$, respectively. then the height of the tower is closest to:

Solution

Height of tower = OP

AB = 42 m

$$tan45\degree = \frac{OP}{OA}$$

1 = OP/OA

OP = OA

$$tan60\degree = \frac{OP}{OA + AB}$$

$$\sqrt{3} = \frac{OP}{OP+ 42}$$

$$\sqrt{3} \times$$ (OP + 42) = OP

OP= $$\frac{42\sqrt{3}}{0.732} = \frac{72.744}{0.732}$$= 99.37 = 99.4 m

The height of tower is 99.4 m.

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