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  • 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 P between them on the road, the angle of elevation of the top of a pole is 60o and the angle of depression from the to

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 P between them on the road, the angle of elevation of the top of a pole is 60o and the angle of depression from the top of the other pole of point P is 30o. Find the heights of the poles and the distance of the point P from the poles.

 

 
 
 
 
 

Answers (1)

Given length of road = 80m, So, BC = 80m.

Let two poles be AB and CD, So, length of pole = AB =CD

P is the point on the road between two poles.

We need to find the height of poles i.e AB and CD and distance of the point (P) from each pole i.e BP and CP

In a right angle triangle \Delta DPC

            \tan P = \frac{CD}{CP}

            CD = \frac{CP}{\sqrt3}\qquad -(i)

Now in right angle triangle \Delta APB

            \tan P = \frac{AB}{PB}

            \tan 60\degree = \frac{AB}{PB}

     \Rightarrow AB = \sqrt{3}BP

    \Rightarrow CD = \sqrt{3}BP\quad [\because AB = CD] \qquad -(ii)

From equation (i) and (ii)

                \frac{CP}{\sqrt{3}} = \sqrt{3}BP \Rightarrow CP = 3BP

Now,

                \\BC = BP + CP \\80 = BP + CP \\80 = BP + 3 BP

                \\\Rightarrow 4BP = 80 \\\Rightarrow BP = 20\qquad \underline{\text{Ans}} \\\Rightarrow CP = 3BP = 60\qquad \underline{\text{Ans}}

So, length of Poles AB = CD = \sqrt3BP = 20\sqrt3 m

Ans. Height of poles  = 20\sqrt3 m\qquad \underline{\text{Ans}}

            

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Safeer PP

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