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 45°. Then the height of the tower (in metres) is
A. $$50\sqrt 3 $$
B. $$50$$
C. $$\frac{{50}}{{\sqrt 2 }}$$
D. $$\frac{{50}}{{\sqrt 3 }}$$
Answer: Option B
Solution(By Examveda Team)
Let AB be tower and C is a point on the ground 50 m awayFrom foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {45^ \circ } = \frac{{AB}}{50} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} $$
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