Question

The angle of elevation of top of building from foot of tower is 30 and angle of elevation of top of tower from foot of building is60 if tower is50m then find height of building

Answer 1

Answer:

Step-by-step explanation:

Answer 2

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

$\green{\therefore{\text{Height\:of\:tower=}\frac{50}{3}\:m}}$

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

• In the given question information given about

the angle of elevation of top of building from foot of tower is 30 and angle of elevation of top of tower from foot of building is 60 if tower is50 m.

• We have to Find the height of tower.

$\green{\underline \bold{Given :}} \\ : \implies \text{Angle of elevation of top of a tower from the foot of the building= }60^{\circ} \\ \\ : \implies \text{Angle of elevation of the top of the building to foot of the tower=} 30^{\circ} \\\\ :\implies \text{Height of building= 50 m}\\ \\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Height\:of\:tower= ?}$

• According to given question :

$\text{Let\:Height\:of\:tower\:be\:x}\\\\ \bold{In \: \triangle \: ABC} \\ : \implies tan\:\theta=\frac{\text{perpendicular}}{\text{base}}\\ \\ : \implies tan\:60^{\circ} = \frac{AB}{BC} \\ \\ : \implies \sqrt{3}=\frac{50}{BC}\\ \\ : \implies BC=\frac{50}{\sqrt{3}}\:m\\ \\ \bold{In\:\triangle\:DCB}\\ :\implies tan\:\theta=\frac{p}{b} \\\\ :\implies tan\:30^{\circ}=\frac{DC}{BC}\\\\ :\implies \frac{1}{\sqrt{3}}=\frac{\sqrt{3}DC}{50}\\\\ \green{:\implies DC=\frac{50}{3}}\\\\ \green{\therefore \text{height \: of \: tower =}\frac{50}{3}\:m}$