The same question draw the diagram for it's journey

Here is a description of the diagram for the aeroplane's journey: Step 1: Mark the starting point. Draw a point and label it A (Ikeja airport). Step 2: Draw the first leg of the journey. From point A, draw a vertical line segment pointing upwards (representing North). Label the end of this segment B. Write "500 km" along this segment. This represents the aeroplane flying due North from A to B. Step 3: Draw the North line at point B. From point B, draw another vertical line segment pointing upwards (parallel to the first North line from A). This is the North direction from B. Step 4: Draw the second leg of the journey. From point B, measure an angle of 60^ clockwise from the North line you just drew at B. Draw a line segment in this direction. Label the end of this segment C (the road junction). Write "300 km" along this segment. This represents the aeroplane flying on a bearing of 060^ from B to C. Step 5: Connect the final position to the starting point. Draw a straight line segment connecting point A to point C. This represents the direct distance and bearing of the aeroplane from Ikeja airport. Step 6: Label the angles. • The angle between the North line at B (pointing upwards) and the line segment BC is 60^. • The interior angle of the triangle at B, ABC, is 120^. (This is because the line AB is North, so the line BA is South. The angle from South to BC is 180^ - 60^ = 120^). • The angle between the North line at A (the line AB) and the line segment AC is BAC. This is the bearing of C from A, which we calculated as approximately 22^. The diagram would look like this: ` N (North) ^ | | AB = 500 km | B |\ | \ | \ BC = 300 km | \ | C | / | / | / |/ A ` (Imagine the line from B to C is at a 60^ angle clockwise from the vertical North line at B, and the line AC connects A to C. The angle at B inside the triangle is 120^.) That's 2 down. 3 left today — send the next one.