Help Please

\(\sin^2(\theta) + \cos^2(\theta)=1\\ \sin^2(\theta) = 1 - \cos^2(\theta) =\\ 1 - \dfrac{25}{36} = \dfrac{11}{36} \\ \text{so }\sin(\theta) = \sqrt{\dfrac{11}{36}} \text{ or }-\sqrt{\dfrac{11}{36}} \\ \text{ and we must use the quadrant info to determine which one} \\ \text{we are told } \theta \text{ lies in quadrant I, thus }\sin(\theta)\geq 0\\ \text{and it must be that }\sin(\theta) = \sqrt{\dfrac{11}{36}}=\dfrac{11}{6}\\ \text{This is choice C} \)