Included angle from whole-circle bearings: If the whole-circle bearings of lines OA and OB are 16° 10' and 332° 18' respectively, what is the included angle BOA (taken at O from OB to OA, choosing the smaller angle)?

Introduction / Context:
Computing included angles from whole-circle bearings (WCB) is a routine step in traverse adjustment. Careful attention to the direction of subtraction and selection of the smaller angle avoids common sign and wrap-around mistakes.

Given Data / Assumptions:

• Bearing(OA) = 16° 10' (WCB).
• Bearing(OB) = 332° 18' (WCB).
• Included angle BOA is the turn from OB to OA at the station O.

Concept / Approach:
To turn from OB to OA, advance from 332° 18' up to 360°, then continue to 16° 10'. The net turn is (360° − 332° 18') + 16° 10' = 27° 42' + 16° 10' = 43° 52'. Always normalize within 0°–360° and choose the interior (smaller) angle unless otherwise specified.

Step-by-Step Solution:

Verification / Alternative check:
Direct difference |16° 10' − 332° 18'| = 316° 08' (exterior angle); 360° − 316° 08' = 43° 52', confirming the result.

Why Other Options Are Wrong:

• 316° 10' and 348° 08': incorrect arithmetic or wrong direction of subtraction.
• 158° 28': not consistent with either the interior or exterior difference for these bearings.

Common Pitfalls:
Forgetting wrap-around across 0°; mixing quadrantal and whole-circle systems; choosing the exterior angle by mistake.

Final Answer:
43° 52'