The same question draw the diagram for it's journey

Draw a point and label it A (Ikeja airport).

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.

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.

From point B, measure an angle of $60^\circ$ 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 $060^\circ$ from B to C.

Draw a straight line segment connecting point A to point C. This represents the direct distance and bearing of the aeroplane from Ikeja airport.

• The angle between the North line at B (pointing upwards) and the line segment BC is $60^\circ$.
• The interior angle of the triangle at B, $\angle ABC$, is $120^\circ$. (This is because the line AB is North, so the line BA is South. The angle from South to BC is $180^\circ - 60^\circ = 120^\circ$).
• The angle between the North line at A (the line AB) and the line segment AC is $\angle BAC$. This is the bearing of C from A, which we calculated as approximately $22^\circ$.

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 $60^\circ$ 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 $120^\circ$.)