Slide 2:
035 0 N (0 0 ) S E W A A Ship travels from Port A to Port B, 60km away on a bearing of 035 0 Draw a diagram showing this information. If the ship returns to Port A using the same path, determine the bearing from B to A. N (0 0 ) S E W A B B 60km N (0 0 ) S E W 180 0 180 0 + 035 0 = 215 0 035 0 035 0 y
Slide 3:
115 0 N S E W Example1: A Boat leaves port A and sails to port B and then to port C . The bearing of B from A is115 o The bearing of C from B is 035 o The distance AC is 95km and AB is 65km Draw and label the diagram Calculate <ABC Calculate <ACB Bearing A from C Boat leaves port C and travels to Port D . Port D is located due West of Port C and due North of Port A . show Port D on diagram and calculate distance CD. A B N S E W N S E W D C 035 0 25 0 25 0 65 0 90 0 180 0 y k 65km 95km 65 0 + 35 0 = 100 0 42.4 0 257.4 0 x Sine Rule 42.4 35 0 180 0 + 35 0 + 42.4 0 = 257.4 0 90 0 j 35 0 + 42.4 0+ + j = 90 0 j = 12.6 0 12.6 0 Pythagoras Theorem 92.7 km 92.7 km END