Two poles at points A and B stand on a horizontal place. C and D are two points between them. The angle of elevation of the top of the poles as observed from point C  are 30^o, 60^o and from point D are 60^o and 45^o. If CD = 30m,  then AB is equal to

Two vertical poles stand at points A and B on a horizontal line. Points C and D lie between them with CD = 30 m. The angles of elevation of the tops of the poles are:

• From C: 30° and 60°
• From D: 60° and 45°

Find the distance AB.

Configuration

As we move from the point nearer to a pole to farther, its elevation angle decreases. The consistent order on the line is: A — D — C — B.

Notation

• Let AD = x → so AC = x + 30.
• Let CB = y → so DB = 30 + y.
• Let the heights of the poles at A and B be h_A and h_B.

1) Using angles to pole at A

At D: tan 60° = h_A / AD = h_A / x → h_A = √3 · x.

At C: tan 30° = h_A / AC = h_A / (x + 30) = 1 / √3 → h_A = (x + 30) / √3.

Equate: √3 · x = (x + 30)/√3 → 3x = x + 30 → x = 15 m.

2) Using angles to pole at B

At C: tan 60° = h_B / CB = h_B / y → h_B = √3 · y.

At D: tan 45° = h_B / DB = h_B / (30 + y) = 1 → h_B = 30 + y.

Equate: √3 · y = 30 + y → (√3 − 1)y = 30 → y = 30 / (√3 − 1) = 15(√3 + 1) m.

3) Distance between poles

AB = AD + DC + CB = x + 30 + y = 15 + 30 + 15(√3 + 1) = 60 + 15√3 m.

AB = 60 + 15√3 m ≈ 85.98 m (≈ 86 m).