1. **Problem Statement:** Measure bearings and solve related problems involving bearings and distances between points. 2. **Understanding Bearings:** Bearing is the angle measured clockwise from the north direction to the line connecting two points. 3. **Formula:** Bearing = angle clockwise from north to the line AB. --- **Problem 3:** Find bearing of A from B given bearing of B from A = 250°. - Use formula: Bearing of A from B = (bearing of B from A + 180°) mod 360°. - Calculate: (250° + 180°) mod 360° = 430° mod 360° = 70°. **Answer:** Bearing of A from B = $70^\circ$. --- **Problem 4:** Ships A, B, C with bearings and distances. - Place A at origin (0,0). - B at $(10\cos 45^\circ, 10\sin 45^\circ) = (7.07, 7.07)$. - C is east of A, so at $(x,0)$. - Bearing of C from B = 180°, so C lies directly south of B, meaning same x-coordinate as B and y less than B. - Since C is east of A, $y=0$, so $C = (7.07, 0)$. - Distance $AC = \sqrt{(7.07-0)^2 + (0-0)^2} = 7.07$ km. - Vector $CA = A - C = (-7.07, 0)$ points west. - Bearing of A from C is angle clockwise from north to vector CA, which is $270^\circ$. **Answer:** Bearing of A from C = $270^\circ$, distance $AC = 7.07$ km. --- **Problem 5:** Babar Azam walks from P to Q at bearing $50^\circ$ for 3 km, then Q to R at bearing $140^\circ$. Distance PR = 5 km. - Angle between PQ and PR = $140^\circ - 50^\circ = 90^\circ$. - Use law of cosines to find QR: $$QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos 90^\circ = 3^2 + 5^2 - 0 = 9 + 25 = 34$$ - So, $QR = \sqrt{34} \approx 5.83$ km. **Answer:** Distance $QR \approx 5.83$ km. --- **Problem 8:** Car travels 13 km at bearing $40^\circ$. - North component = $13 \times \cos 40^\circ \approx 9.95$ km. - East component = $13 \times \sin 40^\circ \approx 8.36$ km. **Answer:** North = 9.95 km, East = 8.36 km. --- **Summary of final answers:** - Bearing of A from B (Problem 3): $70^\circ$ - Bearing of A from C and distance AC (Problem 4): $270^\circ$, 7.07 km - Distance QR (Problem 5): 5.83 km - North and East components (Problem 8): 9.95 km north, 8.36 km east