Question The angle of elevation of the top Q of a vertical tower PQ from a point x on the ground is 600. From a point y, 40 m vertically above x, the angle of elevation of the top Q of tower is 450. 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 = 60o
Draw line YZ parallel to XP
The angle of elevation of Q from Y, ∠ZYQ = 45o
What is to be found?
PQ and PX
Note: PX = ZY
In right triangle QZY, tan45o = \\frac{QZ}{ZY})
tan 45o = 1
\\frac{QZ}{ZY}) = 1 or QZ = ZY
Let us say ZY = QZ = d
In right triangle QPX, tan60o = \\frac{QP}{PX})
tan 60o = \\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