The bearing of lines OA, OB, OC, and OD are 45°15′, 123°30′,216’45’, and 320°00′ respectively. and the included angles∠AOB ,∠BOC,∠COD and ∠DOA
solution:
∠AOB= Bearing OB – bearing OA
= 123°30′ – 45°15′
= 78°15′
∠ BOC = Bearing of OC – Bearing of OB
= 216°45′ – 123°30′
= 93°15′
∠ COD = Bearing of OD – Bearing of OC
= 320°00′- 216°45′
= 103°15′
∠ DOA = [ (360° – bearing of OD ) + bearing of OA]
= [ (360° – 320°)+ 45°15′ ]
= 85°15′
check:
Sum of the interior angles should be 360°
∠ AOB +∠ BOC +∠ COD +∠ DOA
= 78° 15′ +93°15′ + 103°15′ +85°15′
= 360°