Question

From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower is 60°. If the length of the flag staff is 5 m , them find the height of the tower.​

Answer 1

Refer to the attachment for figure ^^"

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Let , the height of tower (DC) be " x "

In Δ ABC

$\tt \implies \tan(60) = \frac{5 + x}{bc}$

$\tt \implies \sqrt{3} = \frac{5 + x}{bc}$

$\tt \implies bc = \frac{5 + x}{ \sqrt{3} } - - - (i)$

Now , in Δ BDC

$\tt \implies\tan(30) = \frac{x}{bc}$

$\tt \implies \frac{1}{ \sqrt{3} } = \frac{x}{bc}$

$\tt \implies bc = \sqrt{3}x - - - (ii)$

From eq (i) and eq (ii) , we get

$\tt \implies \frac{5 + x}{ \sqrt{3} } = \sqrt{3}x$

$\tt \implies5 + x = 3x$

$\tt \implies2x = 5$

$\tt \implies x = \frac{5}{2}$

$\tt \implies x = 2.5 \: \: m$

Therefore , the height of tower is 2.5 m

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Answer 2

Step-by-step explanation:

Let , the height of tower (DC) be " x "

In Δ ABC

\tt \implies \tan(60) = \frac{5 + x}{bc}⟹tan(60)=bc5+x

\tt \implies \sqrt{3} = \frac{5 + x}{bc}⟹3=bc5+x

\tt \implies bc = \frac{5 + x}{ \sqrt{3} } - - - (i)⟹bc=35+x−−−(i)

Now , in Δ BDC

\tt \implies\tan(30) = \frac{x}{bc}⟹tan(30)=bcx

\tt \implies \frac{1}{ \sqrt{3} } = \frac{x}{bc}⟹31=bcx

\tt \implies bc = \sqrt{3}x - - - (ii)⟹bc=3x−−−(ii)

From eq (i) and eq (ii) , we get

\tt \implies \frac{5 + x}{ \sqrt{3} } = \sqrt{3}x⟹35+x=3x

\tt \implies5 + x = 3x⟹5+x=3x

\tt \implies2x = 5⟹2x=5

\tt \implies x = \frac{5}{2}⟹x=25

\tt \implies x = 2.5 \: \: m⟹x=2.5m

Therefore , the height of tower is 2.5 m

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