two poles of equal heights are standing opposite each other on either side of the road,which is 80m 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 each pole and the distance s of the point from the poles.
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Here is your solution
Given:-
AB and CD be the two poles of equal height.
Their heights be H m.
BC be the 80 m wide road.
P be any point on the road.
Let ,
CP be x m,
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 = 60
Height 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.
hope it helps you
Given:-
AB and CD be the two poles of equal height.
Their heights be H m.
BC be the 80 m wide road.
P be any point on the road.
Let ,
CP be x m,
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 = 60
Height 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.
hope it helps you
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