How do you solve the triangle when a=3, b=2, A=50degrees?

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Answer 1 · 2017-06-23T07:01:35

#hatB~=30.71°#; #hat C~=99.29°#; #c~=3.86#

Since, by the sine theorem, it's

#a/sin hatA=b/sin hatB#

you get

#3/(sin50°)=2/sin hatB#

then

#sin hatB=2/3sin 50°~=0.51#

and #hatB~=30.71°#

Then, since the sum of the angles is #180°#, you get

#hat C=180°-50-30.71~=99.29°#

and you would apply the sine theorem again to find #c#:

#a/sin hatA=c/sin hatC#

#3/(sin50°)=c/(sin 99.29°)#

that's

#c=3(sin99.29°)/(sin50°)~=3.86#