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Question
From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be α and β. Show that the height in miles of the aeroplane above the road is given by `(tan alpha tan beta)/(tan alpha + tan beta)`
Solution
Let h be the height of aeroplane p above the road.
And A and B be the two consecutive milestones, then AB = 1 mile. we have ∠PAQ = α and ∠PBQ = β.
We have to prove
`h = (tan alpha tan beta)/(tan alpha + tan beta)`
The corresponding figure is as follows
In ΔPAQ
`=> tan alpha = (PQ)/(AQ)`
`=> tan alpha = h/x`
`=> x = h/(tan alpha)`
`=> x = h cot alpha`
Again in ΔPBQ
`=> tan beta =(PQ)/(BQ)`
`=> tan beta = h/y`
`=> y = h/(tan beta)`
`=> y = h cot beta`
Now,
`=> AB = x + y`
`=> AB = h(cot alpha + cot beta)`
`=> AB = h(1/tan alpha + 1/tan beta)`
`=> AB = h((tan alpha + tan beta)/(tan alpha tan beta))`
Therefore `h = (tan alpha tan beta)/(tan alpha + tan beta)` (Since A B = 1)
Hence height of aero plane is `(tan alpha tan beta)/(tan alpha + tan beta)`
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