4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

M manish

Let the height of the tower AB is $h$ and the angle of elevation from the ground at point C is $\angle ACB = 30^o$
According to question,
In the right triangle $\Delta ABC$,
$\tan \theta = \frac{AB}{BC} = \frac{h}{30}$
$\tan 30^o =\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

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