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(C) Let $\overline{AB}$ be the tower and $C$ be the object.Given that the distance of the object from the base of the tower is $BC = 30 \ m$.The angle

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$30$

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(C) Let $\overline{AB}$ be the tower and $C$ be the object.Given that the distance of the object from the base of the tower is $BC = 30 \ m$.The angle of depression from the top $A$ to the object $C$ is $45^\circ$,so $\angle XAC = 45^\circ$.Since the line of sight is parallel to the ground,$\angle ACB = \angle XAC = 45^\circ$ (alternate interior angles).In the right-angled triangle $\Delta ABC$,where $\angle B = 90^\circ$:$	an(C) = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{AB}{BC}$$	an(45^\circ) = \frac{AB}{30}$Since $	an(45^\circ) = 1$,we have:$1 = \frac{AB}{30}$$AB = 30 \ m$.Thus,the height of the tower is $30 \ m$.

Observing from the top of a tower,the angle of depression of an object $30 \ m$ away from the base of the tower is found to be $45^\circ$. Find the height of the tower (in $m$).

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