Maths question- really need help! 10pts to best answer!

I'm having a little trouble with the question below:

A plane flies 11km from an airport on a bearing of 220 degrees, and then flies 15km on a bearing of 340 degrees. How far is the plane from the airport.

I would appreciate any help with this question :) Please explain answer, and walk me through answering question please

If you draw it you will see you have a triangle with sides 11 and 15 with the included angle = 60.
By the cosine rule
c^2 = 11^2+15^2 - 2*11*15cos60
c = 13.45 km

You have to use sin and cos to break each flight into it's N/S and E/W components.

The first part of the flight is 11km at 220°, 220° means flying 50° south of west.
The southward component is 11 km * sin(50°) = 8.43 km S
The westward component is 11 km * cos(50°) = 7.07 km W

340° is 20° west of north.
The northward component is 15 km * cos(20°) = 14.10 km N
The westward component is 15 km * sin(20°) = 5.12 km W

The sum of these two vectors is

14.10 km - 8.43 km = 5.67 km N
7.07 km + 5.12 km = 12.19 km W

To get the distance, add the two with the Pythagorean theorem:

D = sqrt(5.67^2 + 12.19^2) = sqrt(32.1489 + 32.1489) = sqrt(180.745) = 13.44 km

The direction is arctan(5.67 / 12.19) = arctan(0.47) = 24.94° N of W or 294.94°.

I can draw you a diagram if you like. Please PM me.

6

An old trick:

use "northings" and "eastings"

When a plane flies in a certain direction, you can break its movement into components: a north-south component and an east-west component.

The sine and cosine of the course (not bearing) is the factor that determines the proportion of the movement that is attributed to each component.

northings = distance * cos(Course)
eastings = distance * sin(Course)

On the first leg of the trip (course = 220 degrees), the direction is between south and west