Question

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 the poles and the distance of the point from the poles

Answer 1

Here is your solution

instead of % it will be °

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

Answer 2

✿ 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 the above figure, AB = CD,

$\longrightarrow$OB + OD = 80 m

Now,

$\longrightarrow$In right ΔCDO,

$\longrightarrow$tan 30° = CD/OD

$\longrightarrow$1/√3 = CD/OD

$\longrightarrow$CD = OD/√3 … (1)

➠ 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

Substituting 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

Therefore, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively