Question

Two poles of equal heights are standing opposite 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 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.​

Answer 1

Step-by-step explanation:

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Answer 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:

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

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