# Solve the triangle a = 3, b = 2 and alpha = 40 degrees

Solve the triangle
a = 3, b = 2 and alpha = 40 degrees
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dominicsheq8
$a/\mathrm{sin}\left(A\right)=b/\mathrm{sin}\left(B\right)=c/\mathrm{sin}\left(C\right)$
therefore a,b,c are the lengths of the sides
so $3/\mathrm{sin}\left(40deg\right)=2/\mathrm{sin}\left(\theta \right)$ so $\mathrm{arcsin}\left(2\ast \mathrm{sin}\left(40\right)/3\right)=B$ or B=25.37deg
now every triangle is 90deg so 40deg+25.37deg=65.37deg so onlyangle left is 90-65.37=24.63degrees this is opposite of lengthc
so law of sines again using a and c now $a/\mathrm{sin}\left(A\right)=c/\mathrm{sin}\left(C\right)$so $3/\mathrm{sin}\left(40deg\right)=c/\mathrm{sin}\left(24.63deg\right)$ solve for c=1.945 units in length
the triangle's properties are...
a=3 b=4 c=1.945 A=40deg B=25.37deg C=24.63deg