Question

A vertical tower stands on a declivity which is, inclined at ${15}^{\circ }$ to the horizon. From the foot of the tower a man ascends the declivity for 80 feet and then finds that the tower subtends an angle of ${30}^{\circ }$. Prove that the height of the tower is
$40(\sqrt{6}-\sqrt{2})$ feet.

Answer 1

Using sine formula in $△ABC$:
$\frac{BC}{\sin {75}^{\circ }}=\frac{AB}{\sin {30}^{\circ }}$

GIven that: $BC=80ft$

$AB=h=80\frac{\sin {30}^{\circ }}{\sin ({45}^{\circ }+{30}^{\circ })}=\frac{40\times 2\sqrt{2}}{[\sqrt{\left(3\right)}+1]}ft.$

$=\frac{40.2\sqrt{2}(\sqrt{3}-1)}{3-1}ft.=40\sqrt{2}(\sqrt{3}-1)ft.=40(\sqrt{6}-\sqrt{2})ft.$.