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. $108$
• B. $54\sqrt{3}$
• C. $36\sqrt{3}$
• D. $54$

Answer 1

The correct option is D $54$

According to the question the figure is

Now from the figure
$OA=hOB=\sqrt{3}hO{B}^2=O{A}^2+A{B}^{2}3h^2=h^2+(54\sqrt{2}{)}^2\Rightarrow h=54m$