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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
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
