From the top [A] of a vertical wall [AB] of height [30 ~m], the angles of depression of th…

Question

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

Accepted Answer

Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower.

Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles)

and $\angle C P A=\angle T A P=15^{\circ}$

Now, $\tan 60^{\circ}=\frac{A B}{B Q}=\frac{30}{B Q}$

$$
\[\begin{gathered} \Rightarrow  \sqrt{3}=\frac{30}{B Q} \ \Rightarrow  B Q=\frac{30}{\sqrt{3}}=10 \sqrt{3} \end{gathered}$$
\]

$$
\begin{aligned}
& \text { and } \tan 15^{\circ}=\frac{A C}{P C} \\
& \Rightarrow 2-\sqrt{3}=\frac{A C}{10 \sqrt{3}} (\because P C=B Q) \\
& \Rightarrow A C=10 \sqrt{3}(2-\sqrt{3})=20 \sqrt{3}-30 \\
& \therefore B C=A B-A C=30-(20 \sqrt{3}-30)=60-20 \sqrt{3}
\end{aligned}
$$

$\therefore$ Area of quadrilateral $B C P Q$

$$
\begin{aligned}
& =B Q \times B C=10 \sqrt{3} \times(60-20 \sqrt{3}) \\
& =10(60 \sqrt{3}-60) \\
& =600(\sqrt{3}-1) \mathrm{m}^2
\end{aligned}
$$

Solutions · 1