# 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 could try using law of sines or law of cosines to solve for it. – Heavenly96 Apr 9 '17 at 15:45
• There's nothing I can do with sin rule as I don't have any angles and only have 2 sides. Same for cosine rule, I need atleast one angle, and as this is on a non-calculator paper I can't do sin^-1 – Charlie Apr 9 '17 at 15:49

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 $\cos x \lt 0$

• Where are you getting that rule from? – Charlie Apr 9 '17 at 15:40
• @Charlie it is the Pythagorean identity.en.wikipedia.org/wiki/Pythagorean_trigonometric_identity – Heavenly96 Apr 9 '17 at 15:44
• Does $sin² x$ just mean $sin x²$ ? – Charlie Apr 9 '17 at 15:50
• @Charlie: $\sin^2 x$ is the usual way of writing $(\sin x)^2$. I think it is because we often see powers of the trig functions. One would usually read $\sin x^2$ as $\sin (x^2)$ – Ross Millikan Apr 9 '17 at 15:51
• Alright, I see. – Charlie Apr 9 '17 at 15:52

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, $\therefore \cos \ \theta = - \frac {5} {6}$.

• The bit I don't understand is that angle x is obtuse so how can you apply Pythagoras'theorem or the Pythagorean Trig Identity. Surely it isn't a right angle triangle? – Charlie Apr 9 '17 at 18:16
• I know what an obtuse angle is, I'm telling you that I don't understand how you're doing tri Pythagoras theorem on a triangle that isn't right-angled. – Charlie Apr 9 '17 at 19:59