Question

Consider a triangular plot $ABC$ with sides $AB=7m,BC=5m$ and $CA=6m$. A vertical lamp-post at the mid point $D$ of $AC$ subtends an angle ${30}^{\circ }$ at $B$. The height (in $m$) of the lamp-post is :

• A. $\frac{3}{2}\sqrt{21}$
• B. $7\sqrt{3}$
• C. $2\sqrt{21}$
• D. $\frac{2}{3}\sqrt{21}$

Answer 1

The correct option is D $\frac{2}{3}\sqrt{21}$


Length of median $BD=\frac{1}{2}\sqrt{2\left(\right(BC{)}^2+\left(AB{)}^2\right)-(AC{)}^2}$
$=\frac{1}{2}\sqrt{2(25+49)-36}$
$=\sqrt{28}$
$=2\sqrt{7}$
Let $h$ be the height of the tower
So, from $\Delta TBD$
$\because \angle TBD={30}^{\circ }$
$\therefore \tan {30}^{\circ }=\frac{h}{2\sqrt{7}}$
$h=2\sqrt{\frac{7}{3}}=\frac{2}{3}\sqrt{21}m$