I need help on a maths question: A plane flies 300km from an airport, A, on a bearing of 240°.The plane turns and travels for 400km on a bearing of 050°. a) Using a scale of 1 cm to 50km, draw an accurate drawing of the flight path of the plane. b) What is the bearing that the plane must travel on to return to the airport?
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In my answer, I will be assuming the bearing is relative, i.e. we figure out the bearing based on the current direction. I cannot upload a picture so I will be answering part B. Let’s call the starting point A, the second point after 300km B and the third point C. The angle at points A, B and C, will be called a, b and c, respectively. If we draw the diagram, we can see that the point form a triangle ABC, and we know: AB = 300 BC = 400 c = 130 degrees First we can use the Cosine rule to find the length AC. AC^2 = 400^2 + 300^2 + 2*400*300*cos(130) AC^2 = 95730.9736… AC = 309.404 (3dp) Now that we have AC we can use sine rule to find the size of angle c. sin(c) = sin(130) * 300/309.404 c = sin-1(sin(130) * 300/309.404) c = 47.96 degrees (2dp) The bearing from the final direction is a clockwise turn until we are 47.96 degrees from the line BC. So: 180 - 47.96 = 132.04 degrees is the final bearing.
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