Question

Two poles AB and PQ of same height 35 m are standing opposite each other on either side of the road.
The angles of elevation of the top of the poles, from a point C between them on the road, are 60° and
30° respectively. Find the distance between the poles.​

Answer 1

Answer:

20√3 m.

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, the position of the point P is 60 m from C and the height of each pole is 20√3 m.

hope it helps you

Answer 2

See attached figure for the diagram of this question .

Now

Solution

Let

Distance AC = x

and

Distance PC = y

(As shown in the figure attached)

Now

In △ BAC

$tan \: 60 \degree = \dfrac{BA}{AC} \\ \\ \implies \sqrt{3} = \frac{35}{x} \\ \\ \implies\red{ \boxed{\boxed{x = \frac{35}{ \sqrt{3} } m}}}$

Now

In △QPC

$tan \: 30 \degree = \dfrac{QP}{PC} \\ \\ \implies \frac{1}{ \sqrt{ 3} } = \frac{35}{y} \\ \\ \implies\red{ \boxed{\boxed{y = {35}{ \sqrt{3}m } }}}$

So

Distance b/w the poles is

$D = x + y \\ \\ = [ \frac{35}{ \sqrt{3} } + 35 \sqrt{3}] m\\ \\ = [20.2+60.6] m\\ \\ = 80.8m$