Question

A flagstaff standing vertically at the vertex ${}^{\prime }{A}^{\prime }$ of an isosceles triangle $ABC$, subtends angle $\frac{\pi }{3}$ at the midpoint of the base $BC$, and an angle $\frac{\pi }{6}$ at the vertex $B$. If $BC=a$, the height of flagstaff is :

• A. $\frac{a}{2}\sqrt{\frac{3}{2}}$
• B. $\frac{a}{3}\sqrt{\frac{3}{2}}$
• C. $a\sqrt{\frac{3}{2}}$
• D. $\frac{a}{4}\sqrt{\frac{3}{2}}$

Answer 1

The correct option is D $\frac{a}{4}\sqrt{\frac{3}{2}}$

Given,

$\frac{x}{AB}=\tan {30}^{\circ }$

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

$\frac{x}{AD}=\tan {60}^o$ $x\left(AD\sqrt{3}\right)$

$A{B}^2=A{D}^2+B{D}^2$

$3x^2=\frac{x^2}{3}+\frac{a^2}{4}$

$\frac{8x^2}{3}=\frac{a^2}{4}$

$x^2=\frac{3a^2}{32}\Rightarrow x=\frac{a\sqrt{3}}{4\sqrt{2}}$