Question

The angle of elevation of the top of the chimney from a fixed point on the ground is 30 °. On reaching 15 meters, the angle of elevation becomes 60 °. Find the height of the chimney.

Don't Give useless Answer. ​

Answer 1

GIVEN :-

• The angle of elevation of the top of the chimney from a fixed point on the ground is 30°.
• On reaching 15 meters, the angle of elevation becomes 60°.

TO FIND :-

• The height of the chimney.

SOLUTION :-

Let "XY" be the ground and "PQ" be the height of the chimney.

Let CQ = a.

★ In ∆ PCQ,

$\\ : \implies \displaystyle \sf \: \tan \theta = \frac{perpendicular}{base}$

$: \implies \displaystyle \sf \: \tan \theta = \frac{PQ}{CQ}$

$: \implies \displaystyle \sf \: \tan 60 ^{ \circ} = \frac{h}{a}$

$: \implies \displaystyle \sf \: \sqrt{3} = \frac{h}{a} \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \because \tan60 ^{ \circ} = \sqrt{3} \bigg \rgroup$

$: \implies \displaystyle \sf \: \sqrt{3}a = h$

$: \implies \displaystyle \sf \: a = \frac{h}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup Equation \: 1\bigg \rgroup$

★ Now in ∆ PBQ,

$\\ : \implies \displaystyle \sf \: \tan \theta = \frac{perpendicular}{base}$

$: \implies \displaystyle \sf \: \tan \theta = \frac{PQ}{BQ}$

$: \implies \displaystyle \sf \: \tan30 ^{ \circ} = \frac{h}{BC + CQ}$

$: \implies \displaystyle \sf \: \tan30 ^{ \circ} = \frac{h}{15 + a}$

$: \implies \displaystyle \sf \: \frac{1}{ \sqrt{3} } = \frac{h}{15 + a} \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \because \tan30 ^{ \circ} = \dfrac{1}{\sqrt{3}} \bigg \rgroup$

$: \implies \displaystyle \sf \: \sqrt{3} h = 15 + a$

Substitute the value of a from equation 1.

$\\ \\ : \implies \displaystyle \sf \: \sqrt{3} h = 15 + \frac{h}{ \sqrt{3} }$

$: \implies \displaystyle \sf \: \sqrt{3}h - \frac{h}{ \sqrt{3} } = 15$

$: \implies \displaystyle \sf \: h \bigg[ \sqrt{3} - \frac{1}{ \sqrt{3} } \bigg] = 15$

$: \implies \displaystyle \sf \: h \bigg[ \frac{ \sqrt{9} - 1}{ \sqrt{3} } \bigg] = 15$

$: \implies \displaystyle \sf \: h \bigg[ \frac{3- 1}{ \sqrt{3} } \bigg] = 15$

$: \implies \displaystyle \sf \: h \bigg[ \frac{ 2}{ \sqrt{3} } \bigg] = 15$

$: \implies \displaystyle \sf \: 2h = 15 \sqrt{3 }$

$: \implies \underline{ \boxed{\displaystyle \sf \: h = \frac{15 \sqrt{3} }{2} }}$

Answer 2

Answer:

Given :-

• The angle of elevation of the top of the chimney from a fixed point on the ground is 30°. On reaching 15 m, the angle of elevation becomes 60°.

To Find :-

• What is the height of the chimney.

Solution :-

Let, the ground of the chimney be XY and the height is PQ and the altitude 'a' be CQ.

In ∆PCQ,

➙ tan60° = $\dfrac{PQ}{CQ}$

⇒ $\sqrt{3}$ = $\dfrac{h}{a}$

⇒ $\sqrt{3}$a = h

⇒ a = $\dfrac{h}{√3}$

Again in ∆PBQ,

⇒ tan30° = $\dfrac{PQ}{BQ}$

⇒ $\dfrac{1}{\sqrt3}$ = $\dfrac{h}{BC}{CQ}$

⇒ $\dfrac{1}{\sqrt3}$ = $\dfrac{h}{15 + a}$

⇒ $\sqrt{3}$h = 15 + a

According to the question,

⇒ $\sqrt{3}$h = 15 + $\dfrac{h}{\sqrt3}$

⇒ $\sqrt{3}$h - $\dfrac{h}{\sqrt3}$ = 15

⇒ h ( $\sqrt{3}$ - $\dfrac{h}{\sqrt3}$ ) = 15

⇒ h ( $\dfrac{3 - 1}{\sqrt3}$ ) = 15

⇒ h ( $\dfrac{2}{\sqrt3}$ ) = 15

⇒ 2h = ${15\sqrt3}$

➥ h = $\dfrac{15\sqrt3}{2}$

$\therefore$ The height of the chimney is $\boxed{\bold{\small{\dfrac{15\sqrt3}{2}}}}$

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