Question

The angle of elevation of a tower from a station
A due north of it is a and from a station B due
west of A is ß, then the height of the tower is(prove by some different methods)​

Answer 1

EXPLANATION.

Let op be the tower and A is a point due to

north of tower op

Let B be the point due to West of A.

Let h be the height of a tower.

in a right angle triangle OAP and OBP

$\rm \: \to \: \tan( \alpha ) = \frac{h}{oa}$

$\to \: \: \tan( \beta ) = \frac{h}{ob}$

in ∆ OAB we have,

=> OB² = OA² + AB²

=> AB² = OB² - OA²

$\to \: \: ab {}^{2} = h {}^{2} \cot {}^{2} ( \beta ) - h {}^{2} \cot {}^{2} ( \alpha )$

$\to \: ab {}^{2} = h {}^{2} ( \cot {}^{2} ( \beta ) - \cot {}^{2} ( \alpha ) )$

=> AB² = h²[ (cosec²b - 1) - ( cosec²a - 1 ) ]

=> AB² = h² [ cosec²b - cosec²a ]

=> as we know that,

=> cosec ø = 1/ sin ø

$\to \: ab {}^{2} = {h}^{2} ( \frac{ \sin {}^{2} ( \alpha ) - \sin {}^{2} ( \beta ) }{ \sin {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) } )$

$\to \: \: h \: = \frac{ab \sin( \alpha ) \sin( \beta ) }{ \sqrt{ \sin {}^{2} ( \alpha ) - \sin {}^{2} ( \beta ) } }$

Answer 2

Solution :

Let op be the tower let A be the point due north to the tower op and let B be point due west of a such that angle OAP = α and angle OBP = β

let h be the height of be the tower in right angled triangle OAP and BOP we have,

tanα = h/OA and tanβ = h/OB

OA = h cotα and BO = h cotβ

in ΔOAB we have,

=> OB² = OA² + AB²

=> AB² = OB² - OA²

=> AB² = h²cot²β - h²cot²α

=> AB² = h²(cot²β - cot²α)

=> AB² = h² [(cosec²β - 1) - (cosec²α - 1)]

=> AB² = h² (cosec²β) - (cosec²α)

=> AB² = h²[(sin²α - sin²β)/(sin²αsin²β)

=> h = ABsinα sinβ/√sin²α - sin²β