Use the formula given below to find the area $K$ of the triangle specified by $B = 40^{\circ}, C = 70^{\circ}$ , and $b = 6$ .
$K = \frac{b^{2} \sin A \sin C}{2 \sin B}$

Answer

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Answer

$K = \frac{6^{2} \sin 70^{\circ} \sin 70^{\circ}}{2 \sin 40^{\circ}} \approx 7.8063$

Steps

Step 1 ： $A = 180 - B - C = 180 - 40 - 70 = 70^{\circ}$

Step 2 ： $\sin A = \sin 70^{\circ}$

Step 3 ： $K = \frac{6^{2} \sin 70^{\circ} \sin 70^{\circ}}{2 \sin 40^{\circ}} \approx 7.8063$