Question

The angle of elevation of the top of a tower at a point on the ground is 45º. After moving a distance of 100 m towards the foot of the tower, the angle of elevation of the top of the tower is found to be 60º. The height of the tower is

• A. $150\sqrt{3}m$
• B. $(100+50\sqrt{3})m$
• C. $(50+150\sqrt{3})m$
• D. $(150+50\sqrt{3})m$

Answer 1

The correct option is D $(150+50\sqrt{3})m$

Let the height of the tower AB be h.
$In\Delta BAD,$
$\tan {60}^{\circ }=\frac{AB}{AD}$
$\Rightarrow \sqrt{3}=\frac{h}{AD}$
$\Rightarrow AD=\frac{h}{\sqrt{3}}=\frac{\sqrt{3}h}{3}m$ ...(i)
$In\Delta BAC,$
$\tan {45}^{\circ }=\frac{AB}{AC}$
$\Rightarrow 1=\frac{h}{100+\frac{\sqrt{3}h}{3}}$ $[\because AC=CD+AD,\text{and from}(i\left)\right]$
$\Rightarrow h=100+\frac{\sqrt{3}h}{3}$
$\Rightarrow (h-\frac{\sqrt{3}h}{3})=100$
$\Rightarrow h\left(\frac{3-\sqrt{3}}{3}\right)=100$
$\Rightarrow h=\frac{100}{\frac{3-\sqrt{3}}{3}}$
$\Rightarrow h=\frac{300}{3-\sqrt{3}}\frac{3+\sqrt{3}}{3+\sqrt{3}}$
$\Rightarrow h=\frac{300}{6}(3+\sqrt{3})$
$\Rightarrow h=\frac{300}{6}(3+\sqrt{3})$
$\Rightarrow h=(150+50\sqrt{3})m$
Hence, the correct answer is option (d).