1. ## vector

A plane flies 60km due north then makes a 40° turn west and flies another 30 km. The pilots makes a 2nd 40° turn west and lands 50km later. USe vector components to find the planes displacement.
Vector 1: x=0, y = 60
Vector 2: x=30cos40=-22.98, y= 30sin40=19.28
Vector 3: x=50cos40=-38.30, y=50sin40=32.14
Adding all the Xs up I get 61.28 and for y 82.42
To find displacement I do 82.42/61.28=tan theta
tan^-1(82.42/61.28)=52.37
180-42-37=127°
So displacement would be 127°?

2. Originally Posted by iamanoobatmath
Vector 1: x=0, y = 60 sin(90)
Vector 2: x=30cos(130)=-22.98, y= 30sin(130)=19.28
Vector 3: x=50cos(170)=-38.30, y=50sin(170)=32.14
Adding all the Xs up I get 61.28 and for y 82.42
To find displacement I do 82.42/61.28=tan theta
tan^-1(82.42/61.28)=52.37
180-42-37=127°
So displacement would be 127°?

Unfortunately you have mixed up the angle of rotation (changing course), the bearing which the plane is heading and the angle of the components (measured to the positive x-axis).

I've attached a sketch to demonstrate what you should have used instead.

One personal remark: It is a huge difference between sin(2) and sin(2°)

3. What software did you use to produce that image?

4. Originally Posted by iamanoobatmath
What software did you use to produce that image?
I use this program: