1. **Problem Statement:** Measure the bearings of points B from A using a protractor in given diagrams. 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. 4. **Example:** If the bearing of B from A is 045°, it means B lies 45° clockwise from north at A. --- **Problem 2:** Find the bearing of school A from school B. - If bearing of B from A is known, bearing of A from B = (bearing of B from A + 180°) mod 360°. --- **Problem 3:** Two boats A and B are 5 km apart; bearing of B from A is 250°. - Draw line AB with A at origin. - Bearing of B from A = 250° means B is 250° clockwise from north at A. - Bearing of A from B = (250° + 180°) mod 360° = 70°. --- **Problem 4:** Ships A, B, C with given bearings and distances. - Bearing of B from A = 045°. - Ship C is east of A, so bearing of C from A = 090°. - Bearing of C from B = 180°. - Distance AB = 10 km. - To find bearing of A from C and distance AC, use triangle properties and trigonometry. - Using coordinates: - Place A at origin (0,0). - B at (10 cos 45°, 10 sin 45°) = (7.07, 7.07). - C is east of A, so at (x,0). - Since bearing of C from B is 180°, C lies directly south of B. - So, C has same x-coordinate as B, y-coordinate less than B. - Therefore, C = (7.07, y). - Since C is east of A, y=0. - So, C = (7.07, 0). - Distance AC = 7.07 km. - Bearing of A from C is angle from north clockwise to line CA. - Vector CA = A - C = (0 - 7.07, 0 - 0) = (-7.07, 0). - This points west, so bearing = 270°. --- **Problem 5:** Babar Azam walks from P to Q at bearing 050° for 3 km, then Q to R at bearing 140°. - Distance PR = 5 km. - Find QR and bearing of R from P. - Use law of cosines and sines in triangle PQR. - Angle between PQ and PR = 140° - 50° = 90°. - Triangle with sides PQ=3 km, PR=5 km, angle between them 90°. - Using law of cosines: $$QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(90^\circ)$$ $$QR^2 = 3^2 + 5^2 - 0 = 9 + 25 = 34$$ $$QR = \sqrt{34} \approx 5.83 \text{ km}$$ - To find bearing of R from P, use vector addition or law of sines. --- **Problem 6:** Rayyan spots snake north, Sarim 25 m east of Rayyan spots snake. - Bearing of Sarim from snake = 122°. - Find distance Rayyan to snake. - Use triangle with points Rayyan (R), Sarim (S), Snake (Sn). - Given SnS bearing 122°, so angle at snake between north and Sarim is 122°. - Rayyan is north of snake, so distance RS is 25 m east. - Use law of sines or right triangle properties to find distance from Rayyan to snake. --- **Problem 7:** Bilal travels 12 km at bearing 060°, then 5 km at 150°. - Find distance from football ground to home. - Use vector components: - First leg: 12 km at 60° - Second leg: 5 km at 150° - Sum vectors and find resultant distance. --- **Problem 8:** Car travels 13 km at bearing 040°. - Find north and east components: - North = $13 \times \cos 40^\circ \approx 9.95$ km - East = $13 \times \sin 40^\circ \approx 8.36$ km --- **Final answers:** - Bearing of A from B in problem 3: $70^\circ$ - Bearing of A from C in problem 4: $270^\circ$, distance AC = 7.07 km - Distance QR in problem 5: $\sqrt{34} \approx 5.83$ km - North and east components in problem 8: 9.95 km north, 8.36 km east --- This completes the solutions for the given bearing problems.