Verified Solution • 2025–26 Session moderate

QJEEM23Ma7fbc: JEE Main 2023 Mathematics - Class 12 JEEM23Ma7fbc Solution: From the top of a vertical wall of height , the angles of…

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 :

Multiple Choice Options

A $200(3-\sqrt{3})$
B $300(\sqrt{3}-1)$
C $300(\sqrt{3}+1)$
D $600(\sqrt{3}-1)$

Step-by-Step Verified Solution

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}

JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Height and Distance Question 1 English Explanation

Now,

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

\begin{array}{ll}&amp;\Rightarrow \sqrt{3}=\frac{30}{B Q} \\\\&amp;\Rightarrow B Q=\frac{30}{\sqrt{3}}=10 \sqrt{3}\end{array}

\begin{aligned}&amp; \text { and } \tan 15^{\circ}=\frac{A C}{P C} \\\\&amp; \Rightarrow 2-\sqrt{3}=\frac{A C}{10 \sqrt{3}} (\because P C=B Q) \\\\&amp; \Rightarrow A C=10 \sqrt{3}(2-\sqrt{3})=20 \sqrt{3}-30 \\\\&amp; \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}&amp; =B Q \times B C=10 \sqrt{3} \times(60-20 \sqrt{3}) \\\\&amp; =10(60 \sqrt{3}-60) \\\\&amp; =600(\sqrt{3}-1) \mathrm{m}^2\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring