An airplane is flying on a compass heading (bearing) of 320 at 335 mph. A wind is blowing with the bearing 300degrees at 50 mph. (a) Find the component form of the velocity of the airplane. (b) Find the actual ground speed and direction of the plane.

Explanation:

In navigation the angle of the course (on a compass) is counted clockwise from the North (so, the direction to the North is  0°, to the East is 90°, to the South is 180°, and to the West is 270°).

The North on most maps is a vertically up direction.

In coordinate Geometry and Trigonometry, which we will use, angles are measured counterclockwise from the positive direction of the horizontal X-axis (the East on most maps)

Let's make a simple transformation into Trigonometric standard using the direction to the East as an X-axis.:

320° on compass is 90° + ( 360° - 320°) = 130° counterclockwise from the X-axis.

300° on compass is 90° + ( 360° - 300°) = 150° counterclockwise from the X-axis.

This is a problem on addition of two vectors. Each is defined by its amplitude and angle of direction:

airplane (vector  A ) has amplitude  335 mph and angle 130°

wind (vector  W ) has amplitude  50 mph and angle 150°

To add these two vectors, we represent both as sums of X-component and Y-component:

Ax = 345.cos (130°)

Ay = 345.sin (130°)

Wx = 40.cos(150°)

Wy = 40.sin(150°)

Both X-components act along the same direction, both Y-components act along the same direction. So, we can add X-components to get an X-component of the resulting movement and add Y-components to get a Y-component of the resulting movement.

(A+W)x = Ax + Wx = 345.cos(130°) + 40.cos(130°)

(A+W)y = Ay + Wy = 345.sin(150°) + 40.sin(150°)

Knowing two components of the resulting vector of movement, we can easily determine the amplitude |A+W| and direction ∠(A+W)

|A+W| =√(A+W)²x + √(A+W)²y

∠(A+W) = arctan|(A+W)y

                           (A+W)x

seeing the graph would

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