Question

From thee top of a building $60$m high the angle of depression of the top and the bottom of a tower are observed to be ${30}^o$ and ${60}^o$. Find the height of the tower.

• A. $60$m
• B. $40$m
• C. $80$m
• D. $70$m

Answer 1

Answer: (B)

$40$m

in $△ADE$

$\tan {60}^o=\frac{60}{DE}\Rightarrow DE=\frac{60}{\sqrt{3}}=20\sqrt{3}$

and we can see that , $BCDE$ is a rec\tan gle

so, $BC=DE\Rightarrow BC=20\sqrt{3}$

and $BD=CE.......\left(1\right)$

and in $△ABC$

$tan{30}^o=\frac{AB}{20\sqrt{3}}\Rightarrow AB=20\sqrt{3}\times \frac{1}{\sqrt{3}}=20$

Now, as $AD=AB+BD\Rightarrow 60=20+BD\Rightarrow BD=40$

and from (1)

$BD=CE=40$ (which is the height of the Building)

Therefore, Answer is $40$