Question

From the top of a tower, the angles of depression of two objects on the ground on the same side of it, are observed to be 60° and 30° respectively and the distance between the objects is 400 √ m. The height (in m) of the tower is?

Answer 1

Step-by-step explanation:

h=200root6

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Answer 2

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

$\green{\tt{\therefore{Height\:of\:tower=600\:m}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Angle \: of \: depression( \theta) = 60 \degree \\ \\ \tt: \implies Angle \: of \: depression( \theta_{o}) = 30 \degree \\ \\ \tt: \implies Distance \: between \: object = 400 \sqrt{3} \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Height \: of \: tower =?$

• According to given question :

$\tt \circ \:Let \: height \: of \: tower \: be \: x \\ \\ \bold{In \: \triangle \: ABC: } \\ \tt: \implies tan \: \theta= \frac{p}{b} \\ \\ \tt: \implies tan \:60 \degree = \frac{AB}{BC} \\ \\ \tt: \implies \sqrt{3} = \frac{x}{BC} \\ \\ \tt: \implies BC = \frac{x}{ \sqrt{3} } - - - - - (1) \\ \\ \bold{In \: \triangle \: ABD : } \\ \tt: \implies tan \: \theta_{o} = \frac{p}{b} \\ \\ \tt: \implies tan \: 30 \degree = \frac{AB}{BD} \\ \\ \tt: \implies \frac{1}{ \sqrt{3} } = \frac{x}{BC + CD} \\ \\ \tt: \implies \frac{1}{ \sqrt{3}} = \frac{x}{BC+ 400 \sqrt{3} } - - - - - (2) \\ \\ \text{Putting \: value \: of \: BC \: in \: (2)} \\ \\ \tt: \implies \frac{1}{ \sqrt{3} } = \frac{x}{ \frac{x}{ \sqrt{3}} + 400 \sqrt{3} } \\ \\ \tt: \implies \frac{1}{ \sqrt{3} } = \frac{x \sqrt{3} }{x + 1200} \\ \\ \tt: \implies x + 1200 = 3x \\ \\ \tt: \implies 2x = 1200 \\ \\ \tt: \implies x = \frac{1200}{2} \\ \\ \green{\tt: \implies x = 600 \: m} \\ \\ \green{ \tt \therefore Height \: of \: tower \: is \: 600 \: m}$