Question

$P$ and $Q$ are two points observed from the top of a building $10\sqrt{3}\text{m}$ high if the angles of depression of the points are complementary and $PQ=20\text{m}$, then the distance of $P$ from the building is : (P being the farther point)

• A. $30\text{m}$
• B. $75\text{m}$
• C. $45\text{m}$
• D. $40\text{m}$

Answer 1

The correct option is A $30\text{m}$

Angle of depressions are $x$ and $90-x$ as shown in the figure

From, $△POR$

$\tan x=\frac{h}{PR}$

$⟹PR=\frac{h}{\tan x}$ ......(1)

From, $△ROQ$

$\tan (90-x)=\frac{h}{QR}$

$⟹QR=\frac{h}{\tan (90-x)}$

$⟹QR=\frac{h}{\cot x}$ $[\because \tan (90-\theta )=\cot \theta ]$

$⟹QR=h\tan x$ .....(2) $[\because \frac{1}{\cot x}=\tan x]$

$\because PQ=PR-QR$

$⟹20=\frac{h}{\tan x}-h\tan x$

$⟹20\tan x=h-h{\tan }^{2}x$

$⟹20\tan x=10\sqrt{3}-10\sqrt{3}{\tan }^{2}x$ $[\because h=10\sqrt{3}]$

$⟹{\tan }^{2}x+\frac{2}{\sqrt{3}}\tan x-1=0$

$⟹\tan x=\frac{-\frac{2}{\sqrt{3}}\pm \sqrt{{\left(\frac{2}{\sqrt{3}}\right)}^2+4}}{2}$ (using quadratic formula)

$⟹\tan x=\frac{-\frac{2}{\sqrt{3}}\pm \frac{4}{\sqrt{3}}}{2}$

$⟹\tan x=-\frac{1}{\sqrt{3}}\pm \frac{2}{\sqrt{3}}$

$⟹\tan x=\frac{1}{\sqrt{3}}$ or $\tan x=-\sqrt{3}$

Taking only, $\tan x=\frac{1}{\sqrt{3}}$

$⟹x={30}^{\circ }$

From (1) $PR=\frac{10\sqrt{3}}{\tan x}=10\sqrt{3}\times \sqrt{3}=30\text{m}$

Hence, distance of point P is $30\text{m}$