Question

The angle of elevation of a tower as seen from point A and B are 60° and 30° respectively the distance between a b is hundred metres in line with base of tower find the height of a tower

Answer 1

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

${\bold{\therefore Height\:of\:tower=50\sqrt{3}\:m}}$

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

• In the given question information given about the angle of elevation of a tower as seen from point A and B are 60° and 30° respectively the distance between a b is hundred metres in line with base of tower.

• We have to find the height of a tower.

$\underline \bold{Given : } \\ \implies \angle CAD= 60 \degree \\ \\ \implies \angle CBD= 30 \degree \\ \\ \implies AB= 100 \: m \\ \\ \underline \bold{To \: Find: } \\ \implies Height \: of \: tower = ?$

• According to given question :

$\bold{In \: \triangle \: CDB }\\ \implies tan \: 30\degree = \frac{p}{b} \\ \\ \implies \frac{1}{ \sqrt{3} } = \frac{cd}{x + 100} \\ \\ \implies x + 100 = \sqrt{3} CD \\ \\ \implies x = \sqrt{3} CD - 100 \\ \\ \bold{ In \: \triangle \: CDA }\\ \implies tan \: 60 \degree = \frac{p}{b} \\ \\ \implies \sqrt{3} = \frac{CD}{x} \\ \\ \bold{Putting \: value \: of \: x} \\ \implies \sqrt{3} = \frac{CD}{ \sqrt{3}CD- 100 } \\ \\ \implies 3CD- 100 \sqrt{3} = CD \\ \\ \implies 3CD - cd = 100 \sqrt{3} \\ \\ \implies 2CD= 100 \sqrt{3} \\ \\ \implies CD= \frac{100 \sqrt{3} }{2} \\ \\ \implies \bold{CD = 50 \sqrt{3} \: m} \\ \\ \bold{ \therefore Height \: of \: tower \: is \: 50 \sqrt{3} \: m}$

Answer 2

Answer:

$\bold{\therefore height = 50 \sqrt{3} \: m}$

Step-by-step explanation:

□ Given:

• First angle of elevation = 60°

• Second angle of elevation = 30°

• AB = 100m

$\to tan 30 \degree = \frac{p}{100 + x} \\ \\ \to \frac{1}{ \sqrt{ 3} } = \frac{p}{100 + x} \\ \\ \to x = \sqrt{3} p - 100 \\ \\ \bold{again} \\ \to tan \:60 \degree = \frac{p}{x} \\ \\ \to \sqrt{3} = \frac{p}{ \sqrt{3}p - 100 } \\ \\ \to 3p - p = 100 \sqrt{3} \\ \\ \to p = \frac{100 \sqrt{3} }{2} \\ \\ \bold {\to p = 50 \sqrt{3} } \\ \\ \bold{\therefore height = 50 \sqrt{3} \: m}$