Question$\text{Given a triangle with } B = 40^\circ, \, b = 6, \, c = 9.$ $\text{Solve for the angle } C, \text{ the angle } A, \text{ and the side } a.$

Studdy Solution
Solve for $a$:
$a = \frac{6 \cdot \sin 65.5^\circ}{\sin 40^\circ}$
Calculate $a$:
$a = \frac{6 \cdot 0.9063}{0.6428} \approx 8.46$
The angles and side of the triangle are:
$C \approx 74.5^\circ, \quad A \approx 65.5^\circ, \quad a \approx 8.46$