(In advanced, I apologize for not knowing how to make fractions)
Here's the problem:
A triangle has side $c = 8$ and angles $A = \pi/4$ and $B = \pi/6$. Find the length of the side opposite $A$.
Here's where I'm at so far:
Since $A$ is 45 degrees and $B$ is 30 degrees, $C$ must be 105 degrees, which is $7\pi/12$ radians.
The law of sines states: $\sin A / a = \sin B / b = \sin C / c$
So, we can say $\sin(\pi/4) / a = \sin(7\pi/12) / 8$.
Here's where I'm stuck:
This question may seem so novice, but assuming that I'm correct up until this point (please indicate if I'm not), how do I solve this now? The $\sin(7\pi/12)$ is not a value determined from either a 30-60-90 triangle or a 45-45-90 triangle, so I'm not sure what to do with it.
Any and all help is greatly appreciated!
EDIT:
Thanks to Dylan's comment, I got $16\sqrt2 / (\sqrt2 + \sqrt6)$. Silly me though, I forget how to rationalize the denominator… any help in this area is also appreciated.
Another edit: The furthest I got this rationalized so far is $16 / (1 + \sqrt3$) by first multiplying by $sqrt2$ / $sqrt2$. Then, I got $8\sqrt3 – 8$ by multiplying by $1 – \sqrt3$ / $1 – \sqrt3$. Is this correct?
Best Answer
$$ \sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ. $$