In triangle ABC, A=31.4, B=53.7, <C=61.3°, how do you find the area?

1 Answer
Hubert
May 28, 2015

First you can use The Law of Cosines to evaluate the length of the third side C:
C^2=A^2+B^2-2AB cos gamma
C^2=31.4^2+53.7^2-2*31.4*53.7*cos 61.3^@
C^2=985.96+2883.69-3372.36*0.4802
C^2=2250.2427
C=47.44

Now, for the area of the triangle we can use the Heron's formula:
P=sqrt(p(p-a)(p-b)(p-c))
where p=(a+b+c)/2 is half of the circumference.
a=31.4, b=53.7, c=47.44, p=66.27
P=sqrt(66.27*34.87*12.57*18.83)=sqrt(546958.6761)
P=739.57

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