Question

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

Answer 1

If `/_ B = 19^o` and `/_C = 85^o`
then
`/_A = 180^o - (/_B + /_C) = 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(s-a)(s-b)(s-c))`
where `s` is half the perimeter of the triangle

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