Question

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

Answer 1

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`