Math, asked by akhileshawasthi1976, 6 months ago

There is a small island in 200 m wide river and a tall tree stands on the island . P and Q are points directly opposite to each other on the two banks and in the line with the tree . if the angles of elevation of the top of the tree from P and Q are 30 degree and 45 degree respectively , find the height of the tree.​

Answers

Answered by Anonymous
1

Answer:

P and Q are points directly opposite to each other an two banks and in the line with the tree . If the angle of elevation of the top of the tree.

Answered by Anonymous
7

Answer:

Let OA be the Height of the tree 'h' cm.

\qquad\qquad\tiny\dag \: \:\underline{\sf {In \:  \Delta  \: POA : }} \\  \\

:\implies \sf  \tan(30)  =  \dfrac{Opposite \:  side}{ Adjacent  \: side}  \\  \\  \\

:\implies \sf  \tan(30)  =  \dfrac{OA}{OP}  \\  \\  \\

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

:\implies \sf  OP =  \sqrt{3} \: h   \\  \\  \\

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\qquad\qquad\tiny\dag \: \:\underline{\sf {In \:  \Delta  \: QOA : }} \\  \\

:\implies \sf  \tan(45)  =  \dfrac{Opposite \:  side}{ Adjacent  \: side}  \\  \\  \\

:\implies \sf  1=  \dfrac{h}{OQ}  \\  \\  \\

:\implies \sf  OQ = h \\  \\  \\

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\dashrightarrow\:\: \sf PQ = 200  \: m \\  \\  \\

\dashrightarrow\:\: \sf OP + OQ = 200  \: m  \\  \\  \\

\dashrightarrow\:\: \sf  \sqrt{3}  \: h + h = 200  \: m  \\  \\  \\

\dashrightarrow\:\: \sf h \left(\sqrt{3}  + 1 \right)= 200  \: m  \\  \\  \\

\dashrightarrow\:\: \sf h=  \dfrac{200}{ \sqrt{3}  + 1}   \\  \\  \\

\qquad \:  \:  \:  \:  \:  \: \tiny\dag \: \:\underline{\frak{By \:  rationalising \:  denominator \:  we \:  get : }} \\  \\

\dashrightarrow\:\: \sf h=  \dfrac{200}{ \sqrt{3}  + 1}  \times  \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 }   \\  \\  \\

\dashrightarrow\:\: \sf h=  \dfrac{200 \left( \sqrt{3}  - 1 \right)}{ 3   -  1}   \\  \\  \\

\dashrightarrow\:\: \sf h=  \dfrac{200 \left( \sqrt{3}  - 1 \right)}{ 2}   \\  \\  \\

\dashrightarrow\:\: \sf h= 100 \left( \sqrt{3}  - 1 \right)  \\  \\  \\

\qquad \:  \:  \:  \:  \:  \: \tiny\dag \: \:\underline{\frak{By \: taking  \: \sqrt{3} \:  as \:  1.732  \: we  \: get : }} \\  \\

\dashrightarrow\:\: \sf h= 100 \left(1.732  - 1 \right)  \\  \\  \\

\dashrightarrow\:\: \sf h= 100 \left(0.732 \right)  \\  \\  \\

\dashrightarrow\:\: \underline{ \boxed{ \sf h=  73.2 \: m}}\\  \\  \\

\therefore\underline{\textsf{The height of the tree is \textbf{73.2 m}}}. \\

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\begin{array}{| c | c | c | c | c | c |}\cline{1-6} \bf Angles & \bf 0^{o} & \bf 30^{o} & \bf 45^{o} & \bf 60^{o} & \bf 90^{o} \\\cline{1-6} \tt Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}}& \dfrac{\sqrt{3}}{2}& 1\\\cline{1-6} \tt cos \theta & 1 & \dfrac{\sqrt{3}}{2} &\dfrac{1}{\sqrt{2}}&\dfrac{1}{2}&0\\\cline{1-6} \tt tan \theta & 0 & \dfrac{1}{\sqrt{3}} & 1& \sqrt{3} & \infty \\\cline{1-6} \tt cosec \theta & \infty & 2 & \sqrt{2} & \dfrac{2}{\sqrt{3}} &1\\\cline{1-6} \tt sec \theta & 1 & \dfrac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty \\\cline{1-6} \tt cot \theta & \infty & \sqrt{3} &1& \dfrac{1}{\sqrt{3}} &0\\\cline{1-6}\end{array}

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