56.Two peoples of equal heights are standing opposite each other on ei

1.	56.Two peoples of equal heights are standing opposite each other on either side ofthe road, which is 90mwide. From a point between them on the road, the anglesof elevation of the top of the poles are 60 and 30°, respectively. Find the heightof the poles and the distances of the point from the poles.
Answer» Let AB and CD be the two poles of equal height and their heights beH m. BC be the 80 m wide road. P be any point on the road. Let CPbe x m, therefore BP = (80 – x) .Also, ∠APB = 60° and ∠DPC = 30° In right angled triangle DCP,Tan 30° = CD/CP⇒ h/x = 1/√3⇒ h = x/√3 ---------- (1) In right angled triangle ABP,Tan 60° = AB/AP⇒ h/(80 – x) = √3⇒ h = √3(80 – x)⇒ x/√3 = √3(80 – x)⇒ x = 3(80 – x)⇒ x = 240 – 3x⇒ x + 3x = 240⇒ 4x = 240⇒ x = 60Height of the pole, h = x/√3 = 60/√3 = 20√3.Thus, position of the point P is 60 m from C and height of each pole is 20√3 m.please like the solution 👍 ✔️

56.Two peoples of equal heights are standing opposite each other on either side ofthe road, which is 90mwide. From a point between them on the road, the anglesof elevation of the top of the poles are 60 and 30°, respectively. Find the heightof the poles and the distances of the point from the poles.

Answer»

Let AB and CD be the two poles of equal height and their heights beH m. BC be the 80 m wide road. P be any point on the road. Let CPbe x m, therefore BP = (80 – x) .Also, ∠APB = 60° and ∠DPC = 30°

In right angled triangle DCP,Tan 30° = CD/CP⇒ h/x = 1/√3⇒ h = x/√3 ---------- (1)

In right angled triangle ABP,Tan 60° = AB/AP⇒ h/(80 – x) = √3⇒ h = √3(80 – x)⇒ x/√3 = √3(80 – x)⇒ x = 3(80 – x)⇒ x = 240 – 3x⇒ x + 3x = 240⇒ 4x = 240⇒ x = 60Height of the pole, h = x/√3 = 60/√3 = 20√3.Thus, position of the point P is 60 m from C and height of each pole is 20√3 m.please like the solution 👍 ✔️

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