Question

From a point on the ground $40m$ away from the foot of a tower, the angle of elevation of the top of the tower is ${30}^o.$ The angle of elevation of the top of a water tank (on the top of the tower) is ${45}^o.$
Find the height of the tower

Answer 1

Draw a figure based on given instruction:

From figure: $A$ be the point of observer

$BC=$ Tower

$CD=$ Water tank

$\angle BAD={45}^o$, $\angle BAC={30}^o$ and $AB=40m$

From right angle $\Delta ABD$,

$\Rightarrow \tan {45}^o=\frac{BD}{AB}$

$\Rightarrow 1=\frac{BD}{40}$

$\therefore BD=40...\left(1\right)$

From another right angle, $\Delta ABC$

$\Rightarrow \tan {30}^o=\frac{BC}{AB}$

$\Rightarrow 1\sqrt{3}=\frac{BC}{40}$

Or $BC=\frac{40}{\sqrt{3}}$

Or $BC=\frac{40\sqrt{3}}{3}...\left(2\right)$

Now, Height of the tower $=BC=\frac{40\sqrt{3}}{3}=23.1$

Height of the tower is $23.1m$