How do you solve the triangle given #A=116^circ, b=5, c=3#?

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Answer 1 · 2017-02-06T06:03:40

#a=6.87#;
#hatB=40.88°#;
#hatC=23.12°#

By the cosine theorem, it is:

#a^2=b^2+c^2-2bc cos hatA#

Then

#a^2=5^2+3^2-2*5*3*cos116°#

#a^2=25+9-30*(-0.44)=47.15#

Then #a=sqrt(47.15)=6.87#

By the sines theorem, you get:

#a/sin hatA=b/sin hatB#

#->sin hatB=(b*sin hatA)/a#

#=(5*sin116°)/6.87=0.65#

then #hatB=sin^-1 0.65=40.88°#

#hatC=180°-hatA-hatB=180°-116°-40.88°=23.12°#