Question

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)

Answer 1

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.