What are the x and y degree

[Ys*]

I need to calculate the resultant from figure 1. In order to do it I must add the x and y components of the vectors. I thought it was as simple as cos = x components and sin = y components but a solution I saw switched sin and cos base on the degree/direction or a vector.

In the first quadrant calculating sin(25) for the x, and cos(25) for y. In the second quadrant I saw the solution calculate cos(30) for x and sin(30) for y. I don't understand the reasoning.

 

[MarkFL]

The way that second quadrant angle is defined, we can look at some angle sum identities:

\(\displaystyle \sin\left(90^{\circ}+\theta\right)= \sin\left(90^{\circ}\right)\cos(\theta)+\cos\left(90^{\circ}\right)\sin(\theta)= \cos(\theta)\)

\(\displaystyle \cos\left(90^{\circ}+\theta\right)= \cos\left(90^{\circ}\right)\cos(\theta)-\sin\left(90^{\circ}\right)\sin(\theta)= -\sin(\theta)\)

 

[Ys*]

I understand. thank you.