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प्रश्न
The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β. The height of the tower is
पर्याय
\[\frac{d}{cot \alpha + cot \beta}\]
\[\frac{d}{cot \alpha + cot \beta}\]
\[\frac{d}{\tan \beta - \tan \alpha}\]
\[\frac{d}{\tan \beta - \tan \alpha}\]
उत्तर
The given information can be represented with the help of a diagram as below.
Here, CD = h is the height of the tower. Length of BC is taken as x.
In`Δ ACD`
`tan A=(CD)/(AC)`
`tan∝=h/(d+x)`
`h=(d+x)tan∝ `............(1)
In ΔBCD.
`tan ß = CD/BC`
`tan ß=h/x`
`x=h cot ß` ...............(2)
From (1) and (2)
`h=(d+h cot ß)tan ∝`
`h=d tan ∝+h cot ß tan ∝`
`h(1-cot ß tan ∝ )= d tan ∝`
`h=d tan ∝/((1-cot ß tan ∝ ))=d/(cot ∝-cot ß)`
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