Math, asked by ditya329, 3 days ago

solve this problem....please

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Answered by mathdude500
4

Answer:

\qquad\boxed{ \sf{ \: \bf \: Height\:of\:pole =  \dfrac{ \sqrt{3}  + 1}{2} \: m \: }}  \\  \\

Step-by-step explanation:

Let assume that AB be the pole in the park

Further assume that,

BC be the shadow when sun inclination is 45° and BD be the shadow when sun inclination is 30°.

So, According to statement, the length of shadow increases by 1 m when sun elevation changes from 45° to 30°.

\sf\implies  \: DC \:  =  \: 1 \: m \\  \\

Let assume that AB = 'h' m and BC = x m

Now, In right-angle triangle ABC

\sf \: tan45 \degree \:  =  \: \dfrac{AB}{BC}  \\  \\

\sf \: 1 \:  =  \: \dfrac{h}{x}  \\  \\

\bf\implies \: x = h \:  \:  -  -  - (1) \\  \\

Now, In right-angle triangle ABD

\sf \: tan30\degree \:  =  \: \dfrac{AB}{BD}  \\  \\

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

\sf \: \dfrac{1}{ \sqrt{3} }  \:  =  \: \dfrac{h}{1 + h} \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \boxed{ \sf{ \: \because \: x = h }} \\  \\

\sf \:  \sqrt{3}h = 1 + h \\  \\

\sf \:  \sqrt{3}h - h = 1 \\  \\

\sf \:  (\sqrt{3} - 1)h = 1 \\  \\

\sf \: h = \dfrac{1}{ \sqrt{3}  - 1}  \\  \\

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

\sf \: h =  \dfrac{ \sqrt{3}  + 1}{( \sqrt{3})^{2} -  {(1)}^{2}  }  \\  \\

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

\sf\implies \sf \: h =  \dfrac{ \sqrt{3}  + 1}{2} \: m  \\  \\

\sf\implies \bf \: Height\:of\:pole =  \dfrac{ \sqrt{3}  + 1}{2} \: m  \\  \\

\rule{190pt}{2pt}

\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}

Additional Information

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx =  \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx =  \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx}  = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx}  = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x) = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x +  {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x -  {tan}^{2}x = 1  }\\ \\ \bigstar \: \bf{ {cosec}^{2}x -  {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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