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

1 Answer
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#