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प्रश्न
Two pillars of equal lengths stand on either side of a road which is 100 m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are 60° and 30°. Find the length of each pillar and the distance of the point on the road from the pillars. (Use `sqrt3` = 1.732)
उत्तर
Let AB and CD be the two towers of length 'l' m.
Let P be a point in the roadway BD such that BD = 100 m, ∠APB = 60° and ∠CPD = 30°
In ΔABP,
`"AB"/"BP" = tan 60^circ`
`=> "BP" = "l"/(tan 60^circ)`
`=> "BP" = "l"/sqrt3`
In ΔCDP,
`"CD"/"DP" = tan 30^circ`
`=> "CD"/"DP" = 1/sqrt3`
`=> "l"/"DP" = 1/sqrt3`
`=> "DP" = "l" sqrt3`
`=> "DP" = "l" sqrt3`
The total width of the road is the sum of BP and PD:
∴ BP + PD = 100
∴ `l/sqrt3 + l sqrt3 = 100`
∴ `l (1/sqrt3 + sqrt3) = 100`
∴ `l (1/1.732 + 1.732) = 100` ...[Given `sqrt3` = 1.732]
∴ `l ((1 + 1.732 xx 1.732)/1.732) = 100`
∴ `l( (1 + 2.9998)/1.732) = 100`
∴ `l( (3.9998)/1.732) = 100`
∴ `100/2.309 = l`
∴ l = 43.31 m approx.
Hence, length of the piller is 43.31 m
The point is BP = `l/sqrt3`
The length of the pillars (l) and the position of point P (BP).
∴ BP = `l/sqrt3`
∴ BP = `43.31/1.732`
∴ BP = 25.00 meters.
The length of the pillars (l) and the position of point P (PD).
∴ `"DP" = "l" sqrt3`
∴ DP = 43.31 × 1.732
∴ DP = 75.01 meters.
The distance from the first pillar (associated with the 60° angle of elevation) to point P is approximately 25.00 meters.
The distance from the second pillar (associated with the 30° angle of elevation) to point P is approximately 75.01 meters.
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