triangle trig

Law of sines

sin 70^o  / 5.7 = sin x^o / 5       solve for 'x'

Hello, Guest!~

I'm going to elaborate on Electric Pavlov's excellent answer.

So, basically, the Law of Sines says that, according to Khanacademy, a side labeled, "A" over the sine * opposite of it, which is, angle "a", is equal to a side labeled, "B", over the sine * opposite of that side, which is the angle "b".

In short, you end up with, $\frac{Side\ A}{Angle\ a}=\frac{Side\ B}{Angle\ b}$, and you can add the third side the same way: $\frac{Side\ A}{Angle\ a}=\frac{Side\ B}{Angle\ b}=​​\frac{Side\ C}{Angle\ c}$

For your case, we get:

 $\frac{sin(70)}{5.7}=\frac{x}{5} \\ \\ sin(x)*5.7=sin(70)*5 \\ \\sin(x)=\frac{5sin(70)}{5.7} \\ \\x=sin^{-1}(\frac{5sin(70)}{5.7})$

I do hope the math is correct.

From here, you can just plug it in!

I hope you did take away something from my answer!

-TheLovely1