Question

Two lamp-posts AB and CD each of height 100 m are on either side of the road. P is a point on the road between the two lamp-posts. The angles of elevation of the top of the lamp-posts from the point P are 60° and 40°. Find the distance PB and PD.

Answer 1

Given:

AB = CD = 100 m (vertical lamp‑posts).

P lies on the road between the posts.

From P, angle of elevation to top of AB = 40° (left) and to top of CD = 60° (right).

Step-wise calculation:

1. Find PB distance from P to base B of the left post AB.

In right triangle ABP:

`tan 40^circ = (AB)/(PB)`.

So, `PB = (AB)/(tan 40^circ)` 

= `100/(tan 40^circ)` 

= 100 cot 40°

Numeric: tan 40° ≈ 0.8391

⇒ PB ≈ `100/0.8391` ≈ 119.19 m.

Exact form: PB = 100 cot 40°

= 100 tan 50°

2. Find PD distance from P to base D of the right post CD.

In right triangle CDP:

`tan 60^circ = (CD)/(PD)`

So, `PD = (CD)/tan 60^circ` 

= `100/sqrt(3)` 

= `(100sqrt(3))/3`

Numeric: `100/sqrt(3) ≈ 57.735  m` 

⇒ PD ≈ 57.74 m

PB = 100 cot 40° ≈ 119.19 m.

`PD = 100/sqrt(3)` ≈ 57.74 m.