Answer:
[tex]\displaystyle \sin\theta=\frac{12}{13}[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaystyle \cos\theta =-\frac{5}{13}\text{ where $\theta$ is in QII}[/tex]
Recall that cosine is the ratio of the adjacent side to the hypotenuse. Using the Pythagorean Theorem, solve for the opposite side (we can ignore the negative for now):
[tex]o=\sqrt{13^2-5^2}=12[/tex]
And since θ is in QII, sine is positive, and cosine and tangent are both negative.
Sine is the ratio of the opposite side to the hypotenuse. Therefore:
[tex]\displaystyle \sin\theta=\frac{12}{13}[/tex]
