Math, asked by aswinkumar321188, 5 months ago

Two poles of equal heights are standing opposite to each other
on either side of the road, which is 80 m wide. From a point
between them on the road, the angles of elevation of the top of the
poles are 60⁰ and 30⁰ respectively. Find the height of the poles and
distance of the point from the poles.

Answers

Answered by SANDHIVA1974
2

\large\underline{\underline{\rm{\red{Given:-}}}}

\rm \angle APB = 60^0

\rm \angle DPC = 30^0

\rm BC = 80 \ m

\large\underline{\underline{\rm{\red{Solution:-}}}}

Let BP = x , CP = 80 - x

Now , in \rm \triangle ABP

\rm :\longmapsto tan60^0 = \dfrac{AB}{BP}

\rm :\longmapsto \sqrt3 = \dfrac{AB}{x}

\rm :\longmapsto AB = \sqrt3x

Now in \rm \triangle BPC

\rm :\longmapsto tan30^0 = \dfrac{CD}{CP}

\rm :\longmapsto \dfrac{1}{\sqrt3} = \dfrac{CD}{80 - x}

\rm :\longmapsto CD = \dfrac{80-x}{\sqrt3}

As AB = CD,

\rm :\longmapsto \sqrt3 x = \dfrac{80-x}{\sqrt3}

\rm :\longmapsto 3x = 80 - x

\rm :\longmapsto 4x = 80

\rm :\longmapsto x = 20\ m

Hence, height of pole is given as:

\boxed{\boxed{\rm h = \sqrt3x = 20\sqrt3\ m}}

Distance of point from pole AB

\boxed{\boxed{\rm BP = x \ m = 20 \ m}}

Distance of point from pole CD

\boxed{\boxed{\rm CP = 80 - x = 60 \ m}}

More to know:

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

\color{#FF7968}{\rule{36pt}{7pt}}\color{#4FB3F6}{\rule{36pt}{7pt}}\color{#FBBE2E}{\rule{36pt}{7pt}}\color{#60D399}{\rule{36pt}{7pt}}\color{#6D83F3}{\rule{36pt}{7pt}}

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