From the top A of a vertical wall AB of height 30m, the angles of depression of the top P …

Question

From the top $A$ of a vertical wall $AB$ of height 30m, the angles of depression of the top $P$ and bottom $Q$ of a vertical tower $PQ$ are $15^\circ$ and $60^\circ$ respectively, $B$ and $Q$ are on the same horizontal level. If $C$ is a point on $AB$ such that $CB=PQ$, then the area (in m\^2) of the quadrilateral $BCPQ$ is equal to:

Accepted Answer

Let $BQ=x$ (horizontal distance). From $\angle$ of depression $60^\circ$ to $Q$: $\dfrac{AB}{BQ}=\tan60^\circ \implies \dfrac{30}x=\sqrt3 \implies x=10\sqrt3=y$.

From $\angle$ of depression $15^\circ$ to $P$ (at height $CB=PQ$ above $Q$, so vertical drop from $A$ to $P$ is $30-PQ$): $\dfrac{30-PQ}{x}=\tan15^\circ=2-\sqrt3$
$$30-PQ = 10\sqrt3(2-\sqrt3) = 20\sqrt3-30 \implies PQ = 60-20\sqrt3$$

Area of trapezium $BCPQ = CB\times BQ = PQ\times x = (60-20\sqrt3)(10\sqrt3) = 20(3-\sqrt3)\times10\sqrt3$

$$=600\sqrt3(3-\sqrt3)/3\cdots = 600(\sqrt3-1)$$

*(following the source's simplification directly)*

Solutions · 1