The angle of elevation of the top of a vertical tower standing on a horizontal plane is ob…

Question

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45^\circ$ from a point $A$ on the plane. Let $B$ be the point $30\,	ext{m}$ vertically above the point $A$. If the angle of elevation of the top of the tower from $B$ is $30^\circ$, then the distance (in m) of the foot of the tower from the point $A$ is:

Accepted Answer

Let $h$ be the height of the tower and $d$ the distance of its foot from $A$.

$$\tan45^\circ = \frac hd = 1 \implies h=d$$

$$\tan30^\circ = \frac{h-30}d \implies \sqrt3(h-30)=d$$

Substituting $h=d$: $\sqrt3d-30\sqrt3=d \implies d(\sqrt3-1)=30\sqrt3 \implies d = \frac{30\sqrt3}{\sqrt3-1} = 15\sqrt3(\sqrt3+1) = 15(3+\sqrt3)$

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