Question

If the angle of depression from the top of a building to a point situated $10\sqrt{3}m$ away from the bottom of the building is ${60}^o$, then the height of the building will be:

• A. $10m$
• B. $15m$
• C. $30m$
• D. $30\sqrt{3}m$

Answer 1

Answer: (C)

$30m$

R.E.F image

Let C be the point at $10\sqrt{3}m$ away from the foot of the building of height $AB$.

Given the angle of depression = ${60}^{\circ }$

$\therefore \angle C={60}^{\circ }$ {Alternate angle}

$\tan {60}^{\circ }=\frac{AB}{BC}$ $\Rightarrow \sqrt{3}=\frac{AB}{BC}$

$\Rightarrow AB=BC\sqrt{3}$

$\Rightarrow AB=10\sqrt{3}\times \sqrt{3}=30m$