Determining cosine with a sin value in an obtuse triangle Angle $x$ is obtuse and $\sin{x} = \dfrac{\sqrt{11}}{6}$ Work out the value of $\cos{x}$ I've gotten as far as noting that the opposite side is equal to $\sqrt{11}$ and the hypotenuse is equal to $6$. However, I don't know what function to use to find the adjacent as it's not a right-angled triangle because there is an obtuse angle.

You can just use $\sin^2 x + \cos^2 x =1$. That gives you two solutions. The fact that the angle is obtuse says it is in the second quadrant, so $\cos x \lt 0$