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.
On
Answer: $cos \ \theta$ = $- \frac {5}{6}$.
Solution: Square $sin \ \theta$ to get $\frac {11} {36}$. Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, substitute $\frac {11} {36}$ into this equation to get $\frac {11} {36} + \cos^2 \theta = 1$, and subtract from 1, giving us $\cos^2 \theta = \frac {25} {36}$. Taking the square root gives us $\cos \ \theta = \pm \frac {5} {6}$.
To determine which sign we should use, we use the unit circle and see that sine is positive in two quadrants, the first and the second. Since all angles in the first quadrant are positive, that means we are working with the second quadrant (all angles between $\frac {\pi} {2}$ and $\pi$), meaning $cos \ \theta$ is negative in the second quadrant. Hence, we choose the minus sign for the equation, $\therefore \cos \ \theta = - \frac {5} {6}$.
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 $\cos x \lt 0$