An airplane is flying on a compass heading(bearing) of 340 degrees art 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph, how do you find the component form of the velocity of the airplane?

In navigation the angle of the course (on a compass) is counted clockwise from the North (so, the direction to the North is `0^o`, to the East is `90^0`, to the South is `180^o` and to the West is `270^o`).
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.:
`340^o` on a compass is `90^o +(360^o -340^o)=110^o` counterclockwise from the X-axis.
`320^o` on a compass is `90^o +(360^o -320^o)=130^o` 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 `325` (`mph`) and angle `110^o`;
wind (vector `W`) has amplitude `40` (`mph`) and angle `130^o`.

To add these two vectors, we represent both as sums of X-component and Y-component:
`A_X = 345*cos(110^o)`
`A_Y = 345*sin(110^o)`
`W_X = 40*cos(130^o)`
`W_Y = 40*sin(130^0)`

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 = A_X + W_X = 345*cos(110^o)+40*cos(130^o)`
`(A+W)_Y = A_Y + W_Y = 345*sin(110^o)+40*sin(130^o)`

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

`|A+W| = sqrt((A+W)_X^2+(A+W)_Y^2)`

`/_(A+W) = arctan[(A+W)_Y/(A+W)_X]`

We leave the calculations to the person who suggested the problem.

If it's necessary to express the angle `/_(A+W)` in compass format (`/_alpha`) from its standard trigonometric format (`/_beta`), use the conversion formula
`/_alpha = 360^o-(/_beta-90^o) = 450^o-/_beta`
(if the result is greater than `360^o`, subtract `360^o`)