Question

the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° if the building height is 20 mts .find the height of the tower?​

Answer 1

clearly height of tower is 60m

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Height\:of\:tower=60\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° if the building height is 20 mts.

• We have to Find the height of tower.

$\green{\underline \bold{Given :}} \\ : \implies \text{Angle of elevation of top of a tower from the foot of the building= }60^{\circ} \\ \\ : \implies \text{Angle of elevation of the top of the building to foot of tower =} 60^{\circ} \\\\ :\implies \text{Height of building= 20 m}\\ \\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Height\:of\:tower= ?}$

• According to given question :

$\text{Let\:Height\:of\:tower\:be\:x}\\\\ \bold{In \: \triangle \: ABC} \\ : \implies tan\:\theta=\frac{\text{perpendicular}}{\text{base}}\\ \\ : \implies tan\:30^{\circ} = \frac{AB}{BC} \\ \\ : \implies \frac{1}{\sqrt{3}}=\frac{20}{BC}\\ \\ : \implies BC=20\sqrt{3}\:m\\ \\ \bold{In\:\triangle\:DCB}\\ :\implies tan\:\theta=\frac{p}{b} \\\\ :\implies tan\:60^{\circ}=\frac{DC}{BC}\\\\ :\implies \sqrt{3}=\frac{DC}{20\sqrt{3}}\\\\ \green{:\implies DC=20\sqrt{3}\times\sqrt{3}}\\\\ \green{\therefore \text{height \: of \: tower =60\:m}}$