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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 ..........
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
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