Problem
Use the formula given below to find the area \( K \) of the triangle specified by \( B=40^{\circ}, C=70^{\circ} \), and \( \mathrm{b}=6 \). \[ K=\frac{b^{2} \sin A \sin C}{2 \sin B} \]
Solution
Step 1 ：\( A = 180 - B - C = 180 - 40 - 70 = 70^{\circ} \)
Step 2 ：\( \sin A = \sin 70^{\circ} \)
Step 3 ：\( K = \frac{6^{2} \sin 70^{\circ} \sin 70^{\circ}}{2 \sin 40^{\circ}} \approx 7.8063 \)
