Answer
$$\sin\theta=\frac{\sqrt{11}}{6}$$
Work Step by Step
$$\cos\theta=\frac{5}{6}$$
We can transform the Pythagorean Identities $$\sin^2 x+\cos^2 x=1$$ into $$\sin^2 x=1-\cos^2 x$$
That means $$\sin^2\theta=1-\cos^2\theta=1-\Big(\frac{5}{6}\Big)^2=1-\frac{25}{36}=\frac{11}{36}$$
So, $$\sin\theta=\pm\frac{\sqrt{11}}{6}$$
Another given information shows that $\theta$ is in quadrant I, which means $\sin\theta\gt0$.
Therefore, $$\sin\theta=\frac{\sqrt{11}}{6}$$