Question

If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are ${30}^{\circ }$, ${45}^{\circ }$, and ${60}^{\circ }$ respectively, then the ratio AB:BC is

• A. $1:\sqrt{3}$
• B. $2:3$
• C. $\sqrt{3}:1$
• D. $\sqrt{3}:\sqrt{2}$

Answer 1

The correct option is C $\sqrt{3}:1$



$\because$ PB bisects $\angle APC$, therefore AB : BC = PA : PC
Also $\Delta APQ,sin{30}^{\circ }=\frac{h}{PA}\Rightarrow PA=2h$
And in $\Delta CPQ,sin{60}^{\circ }=\frac{h}{PC}\Rightarrow PA=\frac{2h}{\sqrt{3}}$
$\therefore AB:BC=2h:\frac{2h}{\sqrt{3}}=\sqrt{3}:1$