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Question 71
The horizontal distance between two towers is 60 m. The angular elevation of the top of the taller tower as seen from the top of the shorter one is 30°. If the height of the taller tower is 150 m, the height of the shorter one is ..........
Solution
Let height of shorter tower = CD = AB = x
$$\angle ACP = 30^0$$
AP = PB - AB = 150 - x
AC = BD = 60 m
Now,
$$tan30^0 = \frac{AP}{AC}$$
$$\frac{1}{\sqrt3} = \frac{150 - x}{60}$$
$$150 - x = \frac{60}{\sqrt3} = 20\sqrt3$$
$$x = 150 - 20\sqrt3$$
x = 150 - 34.64
x = 115.36 = 116 m
