Q. 123.9( 11 Votes )
Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 meters, find the distance between the two men. [Take √3 = 1.732.]
Let the positions of the two men are B and C. Let AP represent the tower in the figure. Join B, P, and C, P.We have, ∠ABP = 30° and ∠ACP = 45°. Given that AP = 50 m.We get two right-angled triangles ∆APB and ∆APC, both right angled at P. We use trigonometric ratio tan for both the triangles using BP as a base and AP as height(for ∆APB) and PC as base and AP as height(for ∆APC). We get BP and PC. The distance between the men is BC.
We have to find BC. Let BP = x. In ∆APB,
Again, in ∆APC,
PC = 50
So, distance between the two men is
BC = BP + PC = 50√3 + 50 = 50(√3 + 1) = 136.6m.
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