Q. 313.9( 28 Votes )
From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.
In the above figure, let AB be the tower and P and D be the positions of the two cars at an instant, observed from A. join A and E. The angles of depression are ∠DAE and ∠PAE. Given that, PD = 100 m. Since AE is parallel to BD, so, ∠ADB = ∠DAE = 45° and ∠APB = ∠PAE = 60°. Join P,D and A,B. We get two right-angled triangles ∆ABD and ∆ABP. We are to find AB. We use trigonometric ratio tan for both the triangles, using AB as height and BP as base for ∆ABP and AB as height and BD as base for ∆ABD.
or, AB = BD
or, BD = BP√3
or, BP + 100 = BP√3
or, BP(√3-1) = 100
Hence, the height of the tower is, AB = BD = BP + 100 = 136.61 + 100 = 236.61m
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