Question

. Two poles of equal heights are standing opposite each other on either side of the rol which is 80 m wide. From a point between them on the road, the angles of elevations 1) the top of the poles are 60° and 30°, respectively. Find the height of the poles andal distances of the point from the poles. ​

Answer 1

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

Let AB and CD be two poles of equal height 'h' metres such that AC = 80 m.

Let P be any point on the road AC such that AP = x metres

So, BP = 80 - x metres.

Further given that,

• ∠APB = 60°

• ∠BPC = 30°

Now, In right triangle APB

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{AB}{PB}$

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

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

Now, In right triangle CPD

$\rm :\longmapsto\:tan30 \degree \: = \: \dfrac{CD}{CP}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{h}{80 - x}$

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} } = \dfrac{ \sqrt{3} x}{80 - x}$

$\red{ \bigg\{ \sf \: \because \: using \: equation \: (1) \bigg\}}$

$\rm :\longmapsto\:3x = 80 - x$

$\rm :\longmapsto\:3x + x = 80$

$\rm :\longmapsto\:4x= 80$

$\rm \implies\:\boxed{\tt{ \: x = 20 \: m \: }} - - - (2)$

On substituting the value of x in equation (1), we get

$\rm \implies\:\boxed{\tt{ \: h = 20 \sqrt{3} = 20 \times 1.732 = 34.64 \: m \: }}$

Hence,

• Height of poles, AB and CD = 34.64 m

• Distance of pole AB from P = 20 m

• Distance of pole CD from P = 60 m

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