The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45º. Then the height of the tower (in metres) is

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KollegeApply · Accepted Answer

Suppose AB is the tower and C is a point on the ground. It is given that, BC = 50 m and \angleACB = $45^\circ$. In right ∆ABC, \tan$45^\circ$= AB/BC \Rightarrow 1 = AB/50 \Rightarrow AB = 50 m Thus, the height of the tower is 50 m. [Solution diagram: 63603_solution_1.png]

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