The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation is 45°. Find the height of tower PQ and the distance XQ.

Given,

XY = 40 m

⇒ PQ = h (let)

⇒ ∠PXQ = 60° and ∠RYQ = 45°

⇒ RP = XY = 40 m and RQ = PQ - RP = h - 40

In △RYQ,

$\Rightarrow \tan 45° = \dfrac{RQ}{RY} \\[1em] \Rightarrow 1 = \dfrac{h - 40}{RY} \\[1em] \Rightarrow RY = h - 40.$

From figure,

PX = RY = h - 40

In △PXQ,

$\Rightarrow \tan 60° = \dfrac{PQ}{PX} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{h - 40} \\[1em] \Rightarrow h\sqrt3 - 40\sqrt3 = h \\[1em] \Rightarrow h\sqrt3 - h = 40\sqrt3 \\[1em] \Rightarrow h = \dfrac{40\sqrt3}{(\sqrt3 - 1)} \\[1em] \Rightarrow h = \dfrac{40\sqrt3}{(\sqrt3 - 1)} \times \dfrac{(\sqrt3 + 1)}{(\sqrt3 + 1)} \\[1em] \Rightarrow h = \dfrac{40\sqrt3(\sqrt3 + 1)}{(3 - 1)} \\[1em] \Rightarrow h = \dfrac{40\sqrt3(\sqrt3 + 1)}{2} \\[1em] \Rightarrow h = 20\sqrt3(\sqrt3 + 1) \\[1em] \Rightarrow h = 60 + 20 \times \sqrt3 \\[1em] \Rightarrow h = 60 + 20 \times 1.73 \\[1em] \Rightarrow h = 60 + 34.6 \\[1em] \Rightarrow h = 94.6 \text{ m}$

In △QPX,

$\Rightarrow \sin 60° = \dfrac{PQ}{XQ} \\[1em] \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{PQ}{XQ} \\[1em] \Rightarrow XQ = \dfrac{2PQ}{\sqrt{3}} \\[1em] \Rightarrow XQ = \dfrac{2h}{\sqrt{3}} \\[1em] \Rightarrow XQ = \dfrac{2 \times 94.64}{\sqrt3}\\[1em] \Rightarrow XQ = \dfrac{189.28}{1.732}\\[1em] \Rightarrow XQ = 109.28 \text{ m}.$

Hence, height of tower PQ is 94.64 and XQ = 109.28 m.