Trigonometry-33

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Two pillars are of equal height and on either sides of a road, which is 100 m wide. The angles of elevation of the
top of the pillars are 60⁰ and 30⁰ at a point on the road between the pillars.
Find the position of the point between the pillars and the height of each pillar.

Solution: Let AB and ED be two pillars each of height h meters.

Let C be a point on the road such that BC = x meters.

Then CD = (100 - x) meters.

Given, ∠ACB = 60⁰

Hence, tan 60⁰ = AB / BC

=> √3 = h / x; h = √3x ---------------------------(1)

In triangle ECD, we have tan 30⁰ = ED / CD => 1 / √3 = h / (100 - x)

=> h √3 = 100 - x ----------------------------------------(2)

From (1) and (2) on eliminating h, we have

3x = 100 - x

4x = 100

=> x = 25

Substituting x = 25 in (1), we have h = 25 √3 = 25 X 1.732 = 43.3 m

Thus the required point is at a distance of 25 meters from the first pillar and 75 m from the second pillar.

Hence, the height of each pillar is 43.3 m