# A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is (5pi)/24 and the angle between B and C is  (3pi)/8. What is the area of the triangle?

Jun 15, 2017

33

#### Explanation:

$D e g r e e s = \left(\frac{180}{\pi}\right) \cdot r a \mathrm{di} a n s$
(I just like working in degrees better)

Law of Cosines equation is c^2 = a^2 + b^2 – 2*a*b*cos(gamma).

The angle $\gamma$ needed is 180 – (37.5 + 67.5) = $180 - 105$ = $75$

c^2 = 10^2 + 8^2 – 2*10*8*cos(75).
c^2 = 100 + 64 – 160*(0.259).
${c}^{2} = 122.6$ : $c = 11$
Then using these values we can now find the height $h$ for the triangle and solve for the area. $\sin \left(37.5\right) = \frac{h}{10}$
$h = \sin \left(37.5\right) \cdot 10$ ; $h = 6$
$A = \left(\frac{1}{2}\right) \cdot b \cdot h$ ; $A = \left(\frac{1}{2}\right) \cdot 11 \cdot 6 = 33$