Determine the x and y components of F_1 and F_2.

#### Solution:

Show me the final answers↓

We will start with force F_1.

\sin45^0\,=\,\dfrac{F_{1x}}{200}

(Remember sine is opposite over hypotenuse)

F_{1x}\,=\,200\sin45^0\,=\,141.4 N

\cos45^0\,=\,\dfrac{F_{1y}}{200}

(Remember cosine is adjacent over hypotenuse)

F_{1y}\,=\,200\cos45^0\,=\,141.4 N

Now, we will focus on force F_2. As before, we can write use trigonometry to solve for the components.

\cos30^0\,=\,\dfrac{F_{2x}}{-150}

F_{2x}\,=\,-150\cos30^0\,=\,-130 N

(Notice how our force is negative. This is because the x-component of force F_2 is going towards the negative x-axis.)

\sin30^0\,=\,\dfrac{F_{2y}}{150}

F_{2y}\,=\,150\sin30^0\,=\,75 N

#### Final Answers:

F_{1x}\,=\,141.4 N

F_{1y}\,=\,141.4 N

F_{2x}\,=\,-130 N

F_{2y}\,=\,75 N