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प्रश्न
Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.
उत्तर १
Let AB and CD be the poles and O is the point from where the elevation angles are measured.
In ΔABO
`("AB")/("BO") = tan 60°`
`("AB")/("BO") = sqrt3`
`"BO" = ("AB")/sqrt3`
In ΔCDO,
`("CD")/("DO") = tan 30°`
`("CD")/(80-"BO") = 1/sqrt3`
`"CD"sqrt3 = 80 - "BO"`
`"CD"sqrt3 = 80 - ("AB")/sqrt3`
`"CD"sqrt3+ ("AB")/sqrt3 = 80`
Since the poles are of equal heights,
CD = AB
`"CD"[sqrt3 + 1/sqrt3] = 80`
`"CD"((3+1)/sqrt3) = 80`
`"CD" = 20sqrt3`
`"BO"= ("AB")/sqrt3 = ("CD")/sqrt3`
= `((20sqrt3)/sqrt3)`m
= 20 m
DO = BD − BO
= (80 − 20) m
= 60 m
Therefore, the height of poles is `20sqrt3` and the point is 20 m and 60 m far from these poles.
उत्तर २
Let AB and CD be the two poles of equal height.
O be the point makes an angle of elevation from the top of the poles of 60° and 30°, respectively.
Let OA = 80 - x, OD = x.
And ∠BOA = 30°, ∠COD = 60°.
Here, we have to find the height of the poles and the distance of the points from the poles.
We have the corresponding figure as follows:.
So we use trigonometric ratios.
In a triangle COD,
⇒ tan 60° = `"CD"/"DO"`
⇒ `sqrt(3) = "h"/"x"`
⇒ `"x" = "h"/sqrt(3)`
Again in a triangle AOB,
⇒ tan 30° = `"AB"/"OA"`
⇒ `(1)/sqrt(3) = "h"/(80 -"x")`
⇒ `sqrt(3)"h" = 80 - "x"`
⇒ `sqrt(3)"h" = 80 - "h"/sqrt(3)`
⇒ `sqrt(3)"h" + "h"/sqrt(3) = 80`
⇒ `3"h" + "h" = 80sqrt(3)`
⇒ `4"h" = 80sqrt(3)`
⇒ `"h" = 20sqrt(3)`
⇒ `"x" = (20sqrt(3))/sqrt(3)`
⇒ x = 20
And
⇒ OA = 80 - x
⇒ OA = 80 - 20
⇒ OA = 60
Hence, the height of pole is `20sqrt(3)` and distances are 20m, 60m respectively.
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