Question

The angle of elevation of the top of a vertical tower from a point A due east of it is ${45}^{\circ }$. The angle of elevation of the top of the same tower from a point B, due south of A is ${30}^{\circ }$ If the distance between A and B is $54\sqrt{2}m$,then the height of the tower (in metres),is

• A. $18\sqrt{3}$
• B. $54$
• C. $27\sqrt{2}$
• D. $108$

Answer 1

The correct option is A $18\sqrt{3}$

$\tan 45=\frac{h}{CA}$

$\frac{CA=h}{\sin 45=\frac{h}{DA}}$

$DA=\sqrt{2}h$

$\tan {30}^o=\frac{DA}{AB}$

$\frac{1}{\sqrt{3}}=\frac{\sqrt{2}h}{54\sqrt{2}}$

$h=\frac{54}{\sqrt{3}}=18\sqrt{3}$.