Question

From the top of a cliff $50\, m$ 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. $50\, m$
• 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 $BD =50\, m$ 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. $\triangle DEC$,
$\tan 30^{\circ}=\frac{D C}{C E}=\frac{50-h}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50-h}{x}$
$\Rightarrow x =(50- h ) \sqrt{3}$ ...(i)
In re $\triangle BAD$,
$\tan 45^{\circ}=\frac{D B}{B A}$
$\Rightarrow 1=\frac{50}{x} \Rightarrow x=50$ ...(ii)
$\therefore $ From (i) and (ii)
$50=(50- h ) \sqrt{3}$
$\Rightarrow 50-50 \sqrt{3}=- h \sqrt{3}$
$\Rightarrow h =\frac{50(\sqrt{3}-1)}{\sqrt{3}}$
$=50\left(1-\frac{\sqrt{3}}{3}\right) m .$