Question 78

At a point on level ground, the angle Griensios of a vertical tower is found to be such that its tangent is $$\frac{5}{12}$$. On walking 192 m towards the tower, the tangent of the angle of elevation is $$\frac{3}{4}$$ The height of the tower is ..........

Solution

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BC = 192 m

$$tan\theta = \frac{AP}{AB + BC}$$

$$\frac{5}{12} = \frac{AP}{AB + 192}$$

5AB + 960 = 12AP .......... (1)

$$tan\alpha = \frac{AP}{AB}$$

$$\frac{3}{4} = \frac{AP}{AB}$$

3AB = 4AP

9AB = 12AP .......... (2)

From (1) and (2):

5AB + 960 = 9AB

AB = 240

From equation (2):

9 * 240 = 12AP

AP = 180 m

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