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From the tower 60 m high angles of depre...

From the tower 60 m high angles of depression of the top and bottom of a house are `alphan and beta` respectively. If the height of the house be h. then prove that
`h = (60sin(beta - alpha))/(cos alpha sinbeta)`

Text Solution

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[Hint : In `DeltaABD, tan bet = (60)/(d)`
`rArr " "d = 60 cot beta`………….(1)
In `DeltaDEC, tan alpha =(DC)/(EC)`
`rArr tan alpha = (DC)/(d) `
`rArr DC = d tan alpha ` ltbr`rArr 60 - h = d tan alpha`
`rArr 60-h = 60 cot beta tan alpha` [ by (1) ]
`rArr h= 60(1-(cosbeta)/(sinbeta).(sinalpha)/(cosalpha))`
`rArr h = 60((sinbeta cosalpha - cos beta sinalpha)/(cosalpha sinbeta))`
`rArr h = 60(sin(beta-alpha))/(cos alpha sin beta)` (Proved)]
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