#### Question The angle of elevation of the top Q of a vertical tower PQ from a point x on the ground is 60^{0}. From a point y, 40 m vertically above x, the angle of elevation of the top Q of tower is 45^{0}. Find the height of the tower PQ and the distance PX. (Use \\sqrt3) = 1.73).

#### Video Explanation

#### Explanatory Answer

PQ is the tower.

XY = 40 m

The angle of elevation of Q from X, ∠PXQ = 60^{o}

Draw line YZ parallel to XP

The angle of elevation of Q from Y, ∠ZYQ = 45^{o}

**What is to be found?**

PQ and PX

Note: PX = ZY

In right triangle QZY, tan45^{o} = \\frac{QZ}{ZY})

tan 45^{o} = 1

\\frac{QZ}{ZY}) = 1 or QZ = ZY

Let us say ZY = QZ = d

In right triangle QPX, tan60^{o} = \\frac{QP}{PX})

tan 60^{o} = \\sqrt3)

QP = QZ + ZP = d + 40

PX = d

\\sqrt3) = \\frac{d + 40}{d})

\\sqrt3)d = d + 40

\(\sqrt{3 - 1}))d = 40

Or d = \\frac{40}{\sqrt{3-1}}) = \\frac{40}{\sqrt{3-1}}) × \\frac{\sqrt{3+1}}{\sqrt{3+1}})

= \\frac{40{\sqrt{3+1}}}{3-1}) = \\frac{40{\sqrt{3+1}}}{2})

= 20\\sqrt{3+1}) = 20 (1.73 + 1)

= 20 × 2.73 = 54.6 m

So, d = PX = 54.6m

PQ = d + 40 = 54.6 + 40 = 94.6 m