Math, asked by AestheticSky, 1 month ago

\huge{\underline{\underline{\bf Question}}}

The angle of elevation of a cloud from a point 60m above a lake is 30⁰ and the angle of depression of the reflection of the cloud in the lake is 60⁰. Find the height of the cloud from from the surface of the lake.

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Answers

Answered by GoldenShades
61

\LARGE{\underline{\frak{\orange{Answєr :}}}}

Let AO = H

CD = OB = 60m

A'B = AB = 60 + H

In Δ AOD,

 \tan( {30}^{0} )    =  \frac{ao}{od}  =  \frac{h}{od}  \\  \\ h \:  =  \frac{od}{ \sqrt{} }  \\   \tan( {60}^{0} )  =   \frac{oa}{od}  =  \frac{ob + ba}{od}  \\  \\  \sqrt{3}  =  \frac{60 + 60 + h}{ \sqrt{3h} }   \\  \\ \\  =  \frac{120 + h}{ \sqrt{3} }

⇒ 120 + H = 3H

⇒ 2 H = 120

⇒ H = 60 m

Thus, height of the cloud above the lake = AB + A'B = 60 + 60 = 120 m.

Answered by mathdude500
6

\large\underline{\sf{Given- }}

  • The angle of elevation of a cloud from a point 60m above a lake is 30⁰.

  • The angle of depression of the reflection of the cloud in the lake is 60⁰.

\large\underline{\sf{To\:Find - }}

  • The height of the cloud from from the surface of the lake.

\large\underline{\sf{Solution-}}

  • Let the cloud be at Point A.

Let BC be the lake level and let D be a point 60 meter above the lake level.

  • ⇛ CD = 60 meter

  • ⇛ CD = OB = 60 meter.

  • Let AO = 'h' meter.

Therefore,

  • AB = OA + OB = (60 + h) meter.

Since,

  • Lake level act as a line mirror.

So,

  • Distance of the image = Distance of the object

Therefore,

  • EB = AB = (60 + h) meter.

  • Let OD = 'x' meter.

\rm :\longmapsto\:In \:  \triangle \: AOD

\rm :\longmapsto\:tan30 \degree \: = \dfrac{AO}{OD}

\rm :\longmapsto\: \: \dfrac{1}{ \sqrt{3} }  = \dfrac{h}{x}

 \rm :\implies\:\boxed{ \bf{x \:  =  \:  \sqrt{3} h}} -  -  - (1)

Now,

\rm :\longmapsto\:In  \: \triangle \: DOE

\rm :\longmapsto\:tan60 \degree \: =  \: \dfrac{OE}{DO}

\rm :\longmapsto\: \sqrt{3} \: =  \: \dfrac{OB + BE}{x}

\rm :\longmapsto\: \sqrt{3}  = \dfrac{60 + 60 + h}{ \sqrt{3} h}  \:   \:  \:  \:  \:  \:  \:  \: \{ \because \: of \: (1) \}

\rm :\longmapsto\:3h = 120 + h

\rm :\longmapsto\:2h = 120

\bf\implies  \boxed{ \bf \: \:h \:  =  \: 60 \: m}

Hence,

 \sf \: Height \:  of \:  the \:  cloud \:  from \:  the \:  lake \:  level, \:  =  \: AB

 \sf \: Height \:  of \:  the \:  cloud \:  from \:  the \:  lake \:  level, \:  =  \:60 + h

 \sf \: Height \:  of \:  the \:  cloud \:  from \:  the \:  lake \:  level, \:  =  \: 60 + 60

 \sf \: Height \:  of \:  the \:  cloud \:  from \:  the \:  lake \:  level, \:  =  \: 120 \: meter.

Additional Information :-

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

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