Am I right, what's your opinion? trigonometry

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1MileCrash
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a=4, b=1, C=120*

do you agree, that if you do not round until the final amswer, that angle B = 10.90? rounded to the nearest hundredth?

correct answer given is 10.92...
 
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Thats what i get too, actually. I just remembered it being 89 and wrote 90 out of memory.

I think the difference is that i used the square root of 21 throughout all calculations rather than 4.38 or whatever it roughly is.
 
I don't get either of those! I get B= 10.98 degrees.

I assume that you have a triangle in standard notation- a is the side opposite angle A, etc. Since C is the angle between sides a and b, we must first use the "cosine law": [itex]c^2= a^2+ b^2- 2abCos C[/itex]
[itex]c^2= 4^2+ 1^2- 2(4)(1)(-0.5)= 16+ 1+ 4= 21[/itex]
[itex]c= \sqrt{21}= 4.5826[/itex]

Then, by the sine law,
[tex]\frac{sin(B)}{b}= \frac{sin(C)}{c}[/tex]
[tex]\frac{sin(B)}{1}= \frac{sin(120)}{4.5826}= \frac{.8660}{4.5826}= 0.18897[/tex]
so that [tex]B= 10.89.<br /> <br /> (Oops, a little late!)[/tex]