# 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?

Mar 29, 2018

$2.90$

#### 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 $4 \frac{\pi}{24}$.
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:
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
where $s = \frac{a + b + c}{2}$ or $\frac{\text{perimeter}}{2}$.
$s = \frac{4 + 3 + 2.05}{2} = 4.5$
$A = \sqrt{4.5 \left(4.5 - 4\right) \left(4.5 - 3\right) \left(4.5 - 2.05\right)}$
$A = \sqrt{4.5 \left(0.5\right) \left(1.5\right) \left(2.5\right)} = 2.90$