Question

From the top of a cliff $25m$ height, the angle of the elevation of the the top of tower is found to be equal to the angle of depresion of the foot of the tower. The height of the tower is

• A. $25m$
• B. $50m$
• C. $\frac{25}{\sqrt{2}}m$
• D. $25\sqrt{2}m$

Answer 1

The correct option is B $50m$

R.E.F image

Consider $\Delta DCB$ right angled at c,

$tan{x}^{\circ }=\frac{BC}{CD}=\frac{25}{CD}...\left(1\right)$

Now, consider $\Delta DCE$ right angled atc,

$tan{x}^{\circ }=\frac{CE}{CD}=...\left(2\right)$

equating eq${}^n$ (1) & (2),

$\Rightarrow \frac{25}{CD}=\frac{CE}{CD}$

$\Rightarrow CE=25$

$\therefore$ Total ht. of tower is (25+25) m

($\because$ BE is ht.of tower such that BC+CE & BC = AD = 25 $\Rightarrow 25+CE$ )

$=50m$ Ans

So Option b is correct