Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m 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 the poles and the distances of the point from the poles.
Answers
Answer:
Please mark me as brainleast
Step-by-step explanation:
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
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m 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 the poles and the distances of the point from the poles
Answer
Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken. BD is the distance between the poles.
As per above figure, AB = CD,
OB + OD = 80 m
Now,
In right ΔCDO,
tan 30° = CD/OD
1/√3 = CD/OD
CD = OD/√3 … (1)
Again,
In right ΔABO,
tan 60° = AB/OB
√3 = AB/(80-OD)
AB = √3(80-OD)
AB = CD (Given)
√3(80-OD) = OD/√3 (Using equation (1))
3(80-OD) = OD
240 – 3 OD = OD
4 OD = 240
OD = 60
Putting the value of OD in equation (1)
CD = OD/√3
CD = 60/√3
CD = 20√3 m
Also,
OB + OD = 80 m
⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and
60 m respectively.
