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The angle of elevtion of the top of a to...

The angle of elevtion of the top of a tower from a point A due South of the tower is `alpha` and from a point B due east of the tower is `beta`. If AB = d, then prove that the height of the tower is
`(d)/(sqrt(cot^(2) alpha + cot^(2) beta))`

Text Solution

Verified by Experts

[Hint : Here, OP is the tower whose height is h metres.
In `DeltaPOA, tan alpha = (OP)/(OA)`
`rArr " "OA = OP cot alpha`
`rArr" "OA = h cot alpha`…………..(1)
In `DeltaPOB, tan beta = (OP)/(OB)`
`rArr OB = OP cot beta`
`rArr OB = h cot beta`..............(2)
In ` Delta POB, tan beta = OP/OB `
`rArr d^2 =(h cot alpha )^2 + (h cot B )^2 ` [Form (1) and (2) ]
` rArr d^2 = h^2 ( cot ^2 alpha + cot^2 beta ) `
`rArr h^2 = (d^2)/(cot ^2 alpha + cot ^2 beta )`
` rArr h = ( d) /(sqrt(cos ^2 alpha + cot ^2 beta ))`
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