Math, asked by hiiamauser, 4 days ago

A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point 'A' on the ground is 60 degree and the angle depression of the point 'A' from the top of the tower is 30 degree. Find the height of the tower​

Answers

Answered by mathdude500
2

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

Given that,

  • A pole 6m high is fixed on the top of a tower.

  • The angle of elevation of the top of the pole observed from a point 'A' on the ground is 60 degree.

  • The angle depression of the point 'A' from the top of the tower is 30 degree.

Let assume that BC be the tower of height h and let CD be the pole of 6 m, fixed on the top of tower.

Let AB = x m

Now, ∠CAB = 30° and ∠DAB = 60°.

In right-angle triangle ABC

\rm \: tan30 \degree \:  =  \: \dfrac{BC}{AB}  \\

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

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

Now, In right-angle triangle ABD

\rm \: tan60 \degree \:  =  \: \dfrac{BD}{AB}  \\

\rm \:  \sqrt{3} = \dfrac{6 + h}{x}  \\

On substituting the value of x, we get

\rm \:  \sqrt{3} = \dfrac{6 + h}{ \sqrt{3}  \: h}  \\

\rm \: 3h = 6 + h \\

\rm \: 3h - h = 6 \\

\rm \: 2h = 6 \\

\color{green}\rm\implies \: \: h \:  =  \: 3 \: m \:  \\

So, height of tower, BC = 3 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}

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