Question

From the top of a cliff $50m$ high, the angles of depression of the top and bottom of a tower are observed to be $30^{\circ }$ and $45^{\circ }$. The height of tower is

• A. $50m$
• B. $50\sqrt{3}m$
• C. $50\left(\sqrt{3}-1\right)m$
• D. $50\left(1-\frac{\sqrt{3}}{3}\right)m$

Answer 1

Answer: (D)

Let the height of the cliff $BD=50m$ and the height of the tower $AE=h$ meters .
Given, $\angle DEC=30^{\circ },\angle DAB=45^{\circ }$ Let $BA=CE=x$ meters
In rt. $△DEC$,
$\tan 30^{\circ }=\frac{DC}{CE}=\frac{50-h}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-h}{x}$
$\Rightarrow x=(50-h)\sqrt{3}$ ...(i)
In re $△BAD$,
$\tan 45^{\circ }=\frac{DB}{BA}$
$\Rightarrow 1=\frac{50}{x}\Rightarrow x=50$ ...(ii)
$\therefore$ From (i) and (ii)
$50=(50-h)\sqrt{3}$
$\Rightarrow 50-50\sqrt{3}=-h\sqrt{3}$
$\Rightarrow h=\frac{50(\sqrt{3}-1)}{\sqrt{3}}$
$=50\left(1-\frac{\sqrt{3}}{3}\right)m.$