Trig

Determine the area of a triangle having the following measurements. Round your answer to two decimal places. how would you go about using b=83deg.22" a=6 c=8

$\begin{array}{lcll} B=83^{\circ} 0' 22'' \\ = 83 + \frac{ 0+ \frac{22}{60} } {60} \\ = 83.0061111111^{\circ}\\\\ a=6 \quad c=8 \\\\ \begin{array}{rcl} A&=& ?\\ A &=& \frac12 \cdot a\cdot c \cdot \sin{(B)} \\ A &=& \frac12 \cdot 6\cdot 8 \cdot \sin{(83.0061111111^{\circ})} \\ A &=& \frac12 \cdot 6\cdot 8 \cdot 0.9925591445 \\ \mathbf{A} &\mathbf{=}& \mathbf{23.8214194670} \\ \end{array} \end{array}$

I assume that your triangle is a scalene one, with two known sides and one angle? If that is so, then first you convert the angle to degrees:83 +22/60=83.37 deg. Then you use the Law of Cosines, which is in your Textbook to find the 3rd side. Once you find that, then use Heron Theorem to find the area of your triangle. All these formulas are in your Textbook, if you look hard enough.