ANS:B - 96° 58'

FB of BC = FB of AB+<ABC+-180.
= 152 * 30' + 124 * 28'-180° (first two values sum is more than 180 so use -ve sign).
= 276 * 58'-180° = 96° 58'. Bearing of AB=152°30'.
Angle ABC = 124°28'.
Bearing BC=?
Bearing BA = AB + 180 = 152°30'+180°= 332°30'.
Bearing BC= AB + angle ABC = 332°30' + 124°28' = 456°58'.

Then 456°58'-360°= 96°58'. Fore bearing BC = back bearing AB + clockwise Angle ±180° ±540°.
BC = 152°30'+124°28'±180°±540°.
BC = 276°58'-180°.
BC = 96°58'.
Fore bearing BC=96°58' answer. The reduced bearing of the centerline of two roads AB and AC are N 30° W and N 30° E respectively. These two roads connected by a third road BC. The length and the bearing of BC is 200 m and N 75° E. Find the length of AB and AC.  Bearing of AB=152°30'
<ABC=124°28'

Bearing of BC=?
Back bearing of AB = BA = F.B + 180 = 152°30' + 180° = 332°30'
Bearing of bc = b.b of AB+<ABC = 332°30' + 124°28' = 456°58'

This is more than 360 so;
Bearing of BC = 456°58' - 360° = 96°58'.