The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of tower is β. If AB = d, show that the height of the tower is d/ √ cot ² α+ cot²β

Question

The angle of elevation of the top of a
tower from a point A due south of the tower is α and
from B due east of tower is β If AB = d, show that the height
of the tower is d/ √
cot
2 α+
cot2β

Rayya Halim · Accepted Answer

Let PQ be the tower of height h

and AQ = y, BQ = x

Given: AB = d, ∠PAQ = α, ∠PBQ = β

Now in ∆AQB

AB² = AQ² + BQ² (Pythagoras Theorem)

⇒ d ² = x ²+ y ².......... (1)

In ∆PAQ

$\cot α = \frac{AQ}{PQ} = \frac{y}{h}$

⇒ y = h cot α .............. (2)

In ∆PBQ

$\cot β = \frac{BQ}{PQ} = \frac{x}{h}$

⇒ x = h cot β .............. (3)

Squaring and adding (2) and (3) we get

x ² + y ² = h ²cot² α + h ² cot β

⇒ d ² = h ² (cot² α + cot²β) (from (1))

$\Rightarrow h^{2} = \frac{d^{2}}{\cot^{2} α + \cot^{2} β} \Rightarrow h = \sqrt{\frac{d^{2}}{\cot^{2} α + \cot^{2} β}} = \frac{d}{\sqrt{\cot^{2} α + \cot^{2} β}}$

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The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of tower is β. If AB = d, show that the height of the tower is d/ √ cot 2 α+ cot2β

The angle of elevation of the top of a tower from a point A due south of the tower is α and from B due east of tower is β. If AB = d, show that the height of the tower is d/ √ cot ² α+ cot²β