Two poles of equal height are standing opposite to each other on either side of 150 m wide road .from a point between them on the road ,the angle of elevation of the top of the poles is 60 degree and 30degreee find the height of poles and tye distance of the points from the poles .please answer this question I will mark you as brainliest
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Let AB and CD be the two poles of equal height and their heights be
H m. BC be the 80 m wide road. P be any point on the road. Let CP
be 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 = 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.
Please mark me brainlist
H m. BC be the 80 m wide road. P be any point on the road. Let CP
be 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 = 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.
Please mark me brainlist
Answered by
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this is your right answer
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