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Question
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
Options
\[50\sqrt{3}\]
50
\[\frac{50}{\sqrt{2}}\]
\[\frac{50}{\sqrt{3}}\]
Solution
Suppose AB is the tower and C is a point on the ground.
It is given that, BC = 50 m and \[\angle\]ACB = 45°.
In right ∆ABC,
\[\tan45°= \frac{AB}{BC}\]
\[ \Rightarrow 1 = \frac{AB}{50}\]
\[ \Rightarrow AB = 50 m\]
Thus, the height of the tower is 50 m.
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