two poles of equal height are standing opposite to each other on either side of the road, which is 80m wide. from a point between them on the road, the angle of elevation of the top of the poles are 60° and 30° respectively. find the height of the poles and the distance of the pount from the poles (use √3=1.732)
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Let AB and CD be the poles and O be the point between them on the road.
In triangle ABO
tan 60° = AB/BO
√3 = AB/BO
BO = AB/√3
In triangle COD
tan 30° = CD/DO
1/√3 = CD/DO
DO = CD√3
BO+DO = 80 m {width of road = 80m}
⇒ CD√3 + AB/√3 = 80
(3CD + AB)/√3 = 80
4CD = 80√3 [AB = CD as given height of teo poles is equal]
CD = 20√3 m
CD = AB =34.64 m
Height of the poles = 34.64 m
DO = CD√3 = 20√3*√3 = 60 m
BO+DO = 80 ⇒ BO = 20
The distance of the point from the poles is 20 m, 60 m
In triangle ABO
tan 60° = AB/BO
√3 = AB/BO
BO = AB/√3
In triangle COD
tan 30° = CD/DO
1/√3 = CD/DO
DO = CD√3
BO+DO = 80 m {width of road = 80m}
⇒ CD√3 + AB/√3 = 80
(3CD + AB)/√3 = 80
4CD = 80√3 [AB = CD as given height of teo poles is equal]
CD = 20√3 m
CD = AB =34.64 m
Height of the poles = 34.64 m
DO = CD√3 = 20√3*√3 = 60 m
BO+DO = 80 ⇒ BO = 20
The distance of the point from the poles is 20 m, 60 m
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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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