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(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!


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?

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you have a typo, C must be 105 degrees is $\frac{7\pi{}}{12}$ radians. –  user12205 Sep 14 '11 at 1:43
$7\pi/12 = \pi/4 + \pi/3$, so you could use the addition formula for $\sin$, which would involve computations you seem to be comfortable with. –  Dylan Moreland Sep 14 '11 at 1:45
by the way, you can make fractions using \frac{numerator}{denominator} –  user12205 Sep 14 '11 at 1:49
@Jer: I tried that, but got swept away in a sea of edits! –  The Chaz 2.0 Sep 14 '11 at 1:50
@Mike It doesn't appear that you're having trouble with that at all! –  Dylan Moreland Sep 14 '11 at 2:08

2 Answers 2

up vote 1 down vote accepted

$$ \sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ. $$

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This seems to explain an example of similar kind but with different values .

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What OP has would probably be "ASA", not "AAS"... –  J. M. Sep 14 '11 at 1:51
I think this post can be improved. Would you consider copying and editing the contents of the linked paged to actually answer this question on this site? –  The Chaz 2.0 Sep 14 '11 at 1:51
@J.M i dont think that should me making muh of a difference, i just wanted to make him familiar with since rule.. and regarding actualy trying ot answer this, i am running late for class, so just posted up something that could be of help –  Bhargav Sep 14 '11 at 1:53
Well I appreciate your help 168335 :). However, I've arrived at a rationalizing dillema now. –  Mike Gates Sep 14 '11 at 2:03

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