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do have any reasons why you think it's B?

I used the formula:
\[c ^{2} = a ^{2} + b ^{2} -2ab cosC\]

Couldn't you just use arccos?

hmmm but I want cos C to be by itself
and it's easier to grab the angle A and B first.

Now that I see the formulas you gave me, I think I may be doing the wrong formula.

Fair enough

c is supposed to be the longest side a.k.a hypotenuse

Oh ok. The laws of cosines were getting me confused for a second.

So we would have a = 5.3 b = 4 and c = 7?

so that would mean ... a = 5.3 , b=4, and c = 7...
now we use Cos A to find Angle A

Ok

\[\cos A = \frac{(b^2+c^2-a^2)}{2bc} \]
you just have to plug in those values.. and simplify

Great! Thank you! So we would have cos A = 0.66?

Then in degrees it would be 37.8 degrees?

Can I just double check this for a sec? ^_^ so fast

Oh ok! Yeah go ahead :)

\[\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}\]

Oh nevermind I got 48.7 degrees

I got \[\cos \theta = .6835\]

\[\cos^{-1}(0.66) \rightarrow 48.7\] I also got 48.7 degrees

My bad, I used 54 as the denominator, not 56... ignore the blonde...

ok cool... we got angle A = 48.7 degrees...
Now we need angle B

Oh whoops for the the rest of the equation haha

let's try and solve further.

Ok

\[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]
c = 7, b = 4, a = 5.3

back the equation tool gave me a hard time

I got around 35.1 degrees (34.95 not rounded)

oh wait I got the division wrong
\[\frac{61.09}{74.2} = 0.823 \]

\[\cos^{-1}(0.823) = 34.58\]

|dw:1434937131238:dw|

now use the sin rule

|dw:1434937491614:dw|

With rounding error true

I had angle A almost close to 48.8 as well because when I divided
\[\frac{36.91}{56} \]

if rounding isn't required then it wouldn't be 48.8 but that's cutting it close..