Question

The angle of elevation of top of a tower from the foot of the building is 30° and the angle of elevation of the top of the tower is 45° .if the building is 40m high,find the height of the tower.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

$\green{\therefore{\text{Height\:of\:tower=}\frac{40}{\sqrt{3}}\:m}}$

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

• In the given question information given about the angle of elevation of top of a tower from the foot of the building is 30° and the angle of elevation of the top of the tower is 45° .if the building is 40m high.

• We have to Find the height of tower.

$\green{\underline \bold{Given :}} \\ : \implies \text{Angle of elevation of top of a tower from the foot of the building= }30^{\circ} \\ \\ : \implies \text{Angle of elevation of the top of the tower is =} 45^{\circ} \\\\ :\implies \text{Height of building= 40 m}\\ \\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Height\:of\:tower= ?}$

• According to given question :

$\text{Let\:Height\:of\:tower\:be\:x}\\\\ \bold{In \: \triangle \: ABC} \\ : \implies tan\:\theta=\frac{\text{perpendicular}}{\text{base}}\\ \\ : \implies tan\:45^{\circ} = \frac{AB}{BC} \\ \\ : \implies 1=\frac{40}{BC}\\ \\ : \implies BC=40\:m\\ \\ \bold{In\:\triangle\:DCB}\\ :\implies tan\:\theta=\frac{p}{b} \\\\ :\implies tan\:30^{\circ}=\frac{DC}{BC}\\\\ :\implies \frac{1}{\sqrt{3}}=\frac{DC}{40}\\\\ \green{:\implies DC=\frac{40}{\sqrt{3}}}\\\\ \green{\therefore \text{height \: of \: tower =}\frac{40}{\sqrt{3}}\:m}$