Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is ${30}^{\circ }$. On advancing $150$ meters towards the foot of the tower the angle of elevation increases to ${60}^{\circ }$ . The height of the tower is______

• A. $50(\sqrt{3}+1)$
• B. $50(\sqrt{3}-1)$
• C. $75\left(\sqrt{3}\right)$
• D. None of these

Answer 1

The correct option is C $75\left(\sqrt{3}\right)$

Given

Let $BC=x$

$⟹AC=150+x$

From $\Delta BCD$

$\tan {60}^{\circ }=\frac{CD}{x}$
$\sqrt{3}=\frac{CD}{x}$

$CD=\sqrt{3}x\cdots \left(1\right)$

From $\Delta ACD$

$\tan {30}^{\circ }=\frac{CD}{x+150}$
$\frac{1}{\sqrt{3}}=\frac{CD}{x+150}$

$CD=\frac{1}{\sqrt{3}}\left(x+150\right)\cdots \left(2\right)$

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

$⟹\sqrt{3}x=\frac{1}{\sqrt{3}}(x+150)$

$⟹3x=x+150$
$⟹3x-x=150$

$⟹2x=150$

$⟹x=75$

From $\left(1\right)$

Height of tower is $\sqrt{3}x=75\sqrt{3}$