1. trig help

I have to solve a triangle

the given information in triangle ABC is c=28.6,b=19.9 and A=48.1^o

i used the cosine law to get a which i found to be 21.3 is this correct?

i then tried to use the cosine law again to find one of the other angles

my work:

$
c^2=a^2+b^2=2ab CosC$

$
28^2=21.3^2+19.9^2-2(21.3)(19.9)CosC$

$
CosC=\frac{28.6^2-21.3^2-19.9^2}{-2(21.3)(19.9)}$

$
CosC= \frac{-31.74}{-822.18}$

$CosC=sin^{-1} 0.038$

$C= 87^o$

is this right? i don't think it is because we are working with obtuse triangles

2. Correct, and keep in mind that the law of cosines applies to triangles in general, not just obtuse ones.

3. Originally Posted by william
I have to solve a triangle

the given information in triangle ABC is c=28.6,b=19.9 and A=48.1^o

i used the cosine law to get a which i found to be 21.3 is this correct?

i then tried to use the cosine law again to find one of the other angles

my work:

$
c^2=a^2+b^2=2ab CosC$

$
28^2=21.3^2+19.9^2-2(21.3)(19.9)CosC$

$
CosC=\frac{28.6^2-21.3^2-19.9^2}{-2(21.3)(19.9)}$

$
CosC= \frac{-31.74}{-822.18}$

$CosC=sin^{-1} 0.038$

$C= 87^o$

is this right? i don't think it is because we are working with obtuse triangles
The work you have done appears correct. Now you have the $87^o$ and $48.1^o$ you can simply find the last angle and you're done.