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 BShow Answer
Solution(By Apex Team)
Let AB be tower and C is a point on the ground 50 m away
From foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$\begin{array}{l}\therefore\tan\theta=\frac{AB}{BC}\\
\Rightarrow\tan45^{\circ}=\frac{AB}{50}\\
\Rightarrow1=\frac{AB}{50}\Rightarrow AB=50\mathrm{~m}\end{array}$
From foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$\begin{array}{l}\therefore\tan\theta=\frac{AB}{BC}\\
\Rightarrow\tan45^{\circ}=\frac{AB}{50}\\
\Rightarrow1=\frac{AB}{50}\Rightarrow AB=50\mathrm{~m}\end{array}$
