Question

The angle of elevation of the top of a building from the foot of the tower is and the angle of elevation of the top of the tower from the foot of the building is If the tower is 50 m high, find the height of the building.

Answer 1

Correct Question:

The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

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☯ Let Height of tower and Height of building be CD and AB respectively.

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Given: Height of tower is 50 m.

Now, Let Height of building (AB) be h m.

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$\dag\;{\underline{\frak{In\:right\:angled\: \triangle\: BDC,}}}$

$:\implies\sf tan\:60^\circ = \dfrac{CD}{BD}$

$:\implies\sf \sqrt{3} = \dfrac{50}{BD}\qquad\quad\bigg\lgroup tan\:60^\circ = \sqrt{3} \bigg\rgroup$

$:\implies\sf BD = \dfrac{50}{\sqrt{3}}\:m$

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$\dag\;{\underline{\frak{Now,\:In\:right\:angled\: \triangle\: ABD,}}}$

$:\implies\sf tan\:30^\circ = \dfrac{AB}{BD}$

$:\implies\sf \dfrac{1}{\sqrt{3}} = \dfrac{h}{ \frac{50}{\sqrt{3}}}\qquad\quad\bigg\lgroup tan\:30^\circ = \dfrac{1}{\sqrt{3}} \bigg\rgroup$

$:\implies\sf \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3} h}{50}$

$:\implies\sf h = \dfrac{50}{ \sqrt{3} \times \sqrt{3}}$

$:\implies\sf h = \dfrac{50}{3}$

$:\implies{\underline{\boxed{\frak{\purple{h = 16.66\:m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{The\:height\:of\:the\: buliding\:is\: {\textsf{\textbf{16.66\:m}}}.}}}$

Answer 2

Answer:

Appropriate Question :-

The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

To Find :-

Height

Solution :-

Let the height of tower be CD and height of building be AB

Now,

Height of tower is 50 m.

Let Height of building which we have assumed as AB be h m.

$\tt \implies \tan60 \degree = \dfrac{CD}{BD}$

• As tan60⁰ = √3

$\tt \implies \: \sqrt{3} = \dfrac{50}{BD}$

$\mathfrak \red{BD = \dfrac{50}{ \sqrt{3} } }$

Now, ∆ABD

$\tt \implies \tan30 \degree = \dfrac{AB}{BD}$

• As tan30⁰ = 1/√3

$\tt \implies \dfrac{1} {\sqrt{3} } = \dfrac{ \dfrac{h}{50} }{ \sqrt{3} }$

$\tt \implies \: \dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3h} }{50}$

$\tt \implies \: h = \dfrac{50}{ \sqrt{3} \times \sqrt{3} }$

$\tt \implies \: h = \dfrac{50}{3}$

$\mathfrak \red {h = 18.66 \: m}$

Hence :-

Height of Building is 18.66 m