[tex]\huge{\underline{\underline{\bf Question}}}[/tex]

The angle of elevation of a cloud from a point 60m above a lake

2 thoughts on “[tex]\huge{\underline{\underline{\bf Question}}}[/tex] <br /><br />The angle of elevation of a cloud from a point 60m above a lake”

1. [tex]\LARGE{\underline{\frak{\orange{Answєr :}}}}[/tex]

   Let AO = H

   CD = OB = 60m

   A’B = AB = 60 + H

   In Δ AOD,

   [tex] \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} } [/tex]

   ⇒ 120 + H = 3H

   ⇒ 2 H = 120

   ⇒ H = 60 m

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

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2. [tex]\large\underline{\sf{Given- }}[/tex]

   • 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.

   [tex]\large\underline{\sf{To\:Find – }}[/tex]

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

   [tex]\large\underline{\sf{Solution-}}[/tex]

   • 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.

   [tex]\rm :\longmapsto\:In \: \triangle \: AOD[/tex]

   [tex]\rm :\longmapsto\:tan30 \degree \: = \dfrac{AO}{OD} [/tex]

   [tex]\rm :\longmapsto\: \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{x} [/tex]

   [tex] \rm :\implies\:\boxed{ \bf{x \: = \: \sqrt{3} h}} – – – (1)[/tex]

   Now,

   [tex]\rm :\longmapsto\:In \: \triangle \: DOE[/tex]

   [tex]\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{OE}{DO} [/tex]

   [tex]\rm :\longmapsto\: \sqrt{3} \: = \: \dfrac{OB + BE}{x} [/tex]

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

   [tex]\rm :\longmapsto\:3h = 120 + h[/tex]

   [tex]\rm :\longmapsto\:2h = 120[/tex]

   [tex]\bf\implies \boxed{ \bf \: \:h \: = \: 60 \: m}[/tex]

   Hence,

   [tex] \sf \: Height \: of \: the \: cloud \: from \: the \: lake \: level, \: = \: AB[/tex]

   [tex] \sf \: Height \: of \: the \: cloud \: from \: the \: lake \: level, \: = \:60 + h[/tex]

   [tex] \sf \: Height \: of \: the \: cloud \: from \: the \: lake \: level, \: = \: 60 + 60[/tex]

   [tex] \sf \: Height \: of \: the \: cloud \: from \: the \: lake \: level, \: = \: 120 \: meter.[/tex]

   Additional Information :-

   [tex]\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}[/tex]

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