Question 39

Explanation

Recall the concept of sine addition

Thus

By comparing the coefficients of $\cos \theta$ and $\sin \theta$, we obtain

Square both equations

By adding both equations above, we obtain

To determine the quadrant of $\alpha$, divide equation $\text{(1)}$ by equation $\text{(2)}$

From equation $\text{(1)}$: $r \sin \alpha = 3$. Since $r > 0$, then $\sin \alpha > 0$.

From equation $\text{(2)}$: $r \cos \alpha = -1$. Since $r > 0$, then $\cos \alpha < 0$.

Thus $\alpha$ is in quadrant $\text{II}$.