Question

A triangle has sides A, B, and C. Sides A and B have lengths of 4 and 3, respectively. The angle between A and C is `(11pi)/24` and the angle between B and C is `(3pi)/8`. What is the area of the triangle?

Answer 1

`2.90`

Explanation:

Using the Law of Cosines we can find the other side, and then the area.

Law of Cosines equation is `c^2 = a^2 + b^2 – 2*a*b*cos(gamma)`.
From the given angles, the third angle is `4pi/24`.
`c^2 = 4^2 + 3^2 – 2*4*3*cos(4pi/24)`.
`c^2 = 16 + 9 – 24*0.866 = 4.2`.
`c = 2.05`
We now use Heron's formula for the area:
`A = sqrt(s(s-a)(s-b)(s-c))`
where `s= (a+b+c)/2` or `"perimeter"/2`.
`s = (4+3+2.05)/2 = 4.5`
`A = sqrt(4.5(4.5-4)(4.5-3)(4.5-2.05))`
`A = sqrt(4.5(0.5)(1.5)(2.5)) = 2.90`

http://mste.illinois.edu/dildine/heron/triarea.html
https://www.mathopenref.com/heronsformula.html