Question

the angle of elevation of the top of a building from the foot of a tower is 30° the angle of elevation of the top of the tower from the foot of the building is 60 degree if the tower is 60 M height find the height of the building

Answer 1

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Answer 2

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

$\green{\therefore{\text{Height\:of\:building=20\:m}}}$

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

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

• We have to Find the height of building.

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

• According to given question :

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