Question

The angle of elevation of a tower $CD$ at a place $A$ which is left of it is ${60}^o$ and at a place $B$ due north of $A$, the elevation is ${30}^o$. If $AB=3$m, height of the tower is

• A. $4.5$m
• B. $\frac{2\sqrt{2}}{3\sqrt{3}}$m
• C. $\frac{3\sqrt{3}}{2\sqrt{2}}$m
• D. $\frac{3\sqrt{6}}{2}$m

Answer 1

The correct option is A $4.5$m

Let $CE=x$ m

$AB=DE=3$ m

Height of the tower $=CE+ED=x+3$ m

In $\Delta ACD$

$\tan {60}^o=\frac{CD}{AD}$

$\sqrt{3}=\frac{x+3}{AD}$

$AD=\frac{x+3}{\sqrt{3}}$ ……….$\left(1\right)$

In $\Delta CBE$, $\tan {30}^o=\frac{opposite}{adjacent}=\frac{CF}{BE}$

$\frac{1}{\sqrt{3}}=\frac{x}{AD}$ $[\because BE=AD]$

$AD=x\sqrt{3}$ ……….$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$x\sqrt{3}=\frac{x+3}{\sqrt{3}}$

$3x=x+3$

$3x-x=3$

$2x=3$

$x=\frac{3}{2}$m

Height of tower$=x+3$ m

$=\frac{3}{2}+3=\frac{3+6}{2}$

$=\frac{9}{2}$

$=4.5$m

$\therefore$ Height of tower$=4.5$m.