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प्रश्न
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
उत्तर १
In ΔABC,
`("AB")/("BC")` = tan 60°
`("AB")/("BC") = sqrt3`
`"BC" = ("AB")/sqrt3`
In ΔABD,
`("AB")/("BD") `= tan 30°
`("AB")/("BC"+"CD") = 1/sqrt3`
`("AB")/(("AB")/sqrt3+20) = 1/sqrt3`
`("AB"sqrt3)/("AB"+20sqrt3) = 1/sqrt3`
`3"AB" = "AB"+20sqrt3`
`2"AB" = 20sqrt3`
`"AB" = 10sqrt3 m`
`"BC" = ("AB")/sqrt3`
= `((10sqrt3)/sqrt3)m`
= 10 m
Therefore, the height of the tower is `10sqrt3` m and the width of the canal is 10 m.
उत्तर २
Let PQ = h m be the height of the TV tower and BQ = x m be the width of the canal.
We have,
AB = 20 m, ∠PAQ = 30°, ∠BQ = x and PQ = h
In ΔPBQ,
`tan 60° = ("PQ")/("BQ")`
⇒ `sqrt(3) = h/x`
⇒ `h = x sqrt(3) ` ...(1)
Again in ΔAPQ,
`tan 30° = ("PQ")/("AQ")`
⇒ `1/sqrt(3) = h/("AB" +"BQ")`
⇒ `1/sqrt(3) = (x sqrt(3))/(20+3)` ...[Using (1)]
⇒ 3x = 20 + x
⇒ 3x - x = 20
⇒ 2x = 20
⇒ x = `20/2`
⇒ x = 10 m
Substituting x = 10 in (i), we get
h = `10 sqrt(3)` m
So, the height of the TV tower is 10`sqrt(3)` m and the width of the canal is 10 m.
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