Question

Is anyone available to help with this pre-calc problem? Thanks!

Answer 1

`\begin{gathered} sin\text{ }\theta\text{ = }(-5)/(13) \\ sin\theta\text{ =}(Opposite)/(Hypotenuse) \\ Hypotenuse^2\text{ = Opposite}^2\text{ + Adjacent}^2 \\ Adjacent\text{ = }\sqrt[]{13^2-(-5)^2} \\ =\text{ 12} \\ If\text{ tan}\theta\text{ }>0\text{ and sin}\theta<0 \\ This\text{ means that the angle is in quadrant three where cos is also negative} \\ cos\theta\text{ = }\frac{adjacent\text{ }}{hypotenuse} \\ \text{ = }(-12)/(13)\text{ } \end{gathered}`

Correct option number 3.