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

May 28, 2015

First you can use The Law of Cosines to evaluate the length of the third side $C$:
${C}^{2} = {A}^{2} + {B}^{2} - 2 A B \cos \gamma$
${C}^{2} = {31.4}^{2} + {53.7}^{2} - 2 \cdot 31.4 \cdot 53.7 \cdot \cos {61.3}^{\circ}$
${C}^{2} = 985.96 + 2883.69 - 3372.36 \cdot 0.4802$
${C}^{2} = 2250.2427$
$C = 47.44$

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