# How do you find the area of the triangle given C=85 degrees, a= 2, B= 19 degrees?

May 10, 2015

If $\angle B = {19}^{o}$ and $\angle C = {85}^{o}$
then
$\angle A = {180}^{o} - \left(\angle B + \angle C\right) = {76}^{o}$

We can use the Law of Sines

(2)/(sin(76^o)) = (b)/(sin(19^o)) = (c)/(sin(85^o)

To determine the lengths:
$b = 0.67107$ (approx.)
$c = 2.53384$ (approx.)

Then use Heron's Formula for the area of the triangle
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
where $s$ is half the perimeter of the triangle

In this case (assuming no arithmetic errors)
Area of the triangle$= 0.741216$