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प्रश्न
The angles of elevation of the top of a tower from two points at distance of 5 metres and 20 metres from the base of the tower and is the same straight line with it, are complementary. Find the height of the tower.
उत्तर
Let the height of the tower be AB.
We have.
AC = 5m, AD = 20m
Let the angle of elevation of the top of the tower (i.e. ∠ACB) from point C be θ .
Then,
the angle of elevation of the top of the tower (i.e. Z ADB) from point D
=(90° -θ)
Now, in ΔABC
`tan theta = (AB)/(AC)`
`⇒ tan theta = (AB)/5` ...................(i)
Also, in ΔABD,
`cot (90° - theta ) = (AD)/(AB)`
`⇒ tan theta = 20/(AB)` .................(ii)
From (i) and (ii), we get
`(AB)/5 = 20/(AB)`
`⇒ AB^2 = 100`
`⇒ AB = sqrt(100)`
∴ AB = 10 m
So, the height of the tower is 10 m.
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