This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — here's the solution for part (b): Part (b): Navigation Problem A ship sails from Port A on a bearing 065^ to Port B a distance 54 km. It then sails on the bearing of 155^ from B to Port C a distance of 80 km. Find correct to one decimal place: (i) The distance between A and C (ii) The bearing of C from A. Step 1: Determine the angle ABC. Draw a North line at Port A and Port B. Since these North lines are parallel, we can use properties of parallel lines. The bearing of B from A is 065^. This means the angle between the North line at A and the line segment AB is 65^. The alternate interior angle between the line segment AB and the South line at B is also 65^. The bearing of C from B is 155^. This is the angle measured clockwise from the North line at B to the line segment BC. The angle between the North line at B and the South line at B is 180^. Therefore, the angle between the South line at B and the line segment BC is 180^ - 155^ = 25^. The angle ABC is the sum of the angle between AB and the South line at B, and the angle between the South line at B and BC. ABC = 65^ + 25^ = 90^ Step 2: Calculate the distance AC using the Cosine Rule. In ABC, we have AB = 54 km, BC = 80 km, and ABC = 90^. Since ABC = 90^, we can use the Pythagorean theorem (a special case of the Cosine Rule). AC^2 = AB^2 + BC^2 AC^2 = (54 km)^2 + (80 km)^2 AC^2 = 2916 km^2 + 6400 km^2 AC^2 = 9316 km^2 AC = sqrt(9316) km AC ≈ 96.5194 km Rounding to one decimal place: AC ≈ 96.5 km The distance between A and C is 96.5 km. Step 3: Calculate the bearing of C from A. First, find the angle BAC in ABC. Since it's a right-angled triangle: ( BAC) = OppositeAdjacent = (BC)/(AB) ( BAC) = (80)/(54) BAC = ((80)/(54)) BAC ≈ 55.99^ The bearing of C from A is the sum of the bearing of B from A and the angle BAC. Bearing of C from A = 065^ + BAC Bearing of C from A = 065^ + 55.99^ Bearing of C from A = 120.99^ Rounding to one decimal place: Bearing of C from A ≈ 121.0^ The bearing of C from A is 121.0^. Send me the next one 📸