what is the approximate perimeter of the triangle? use the law of sines to find the answer. law of sines…

Answer:

C. 6.9 units

Explanation:

Step1: Find angle J

The sum of angles in a triangle is 180°. So, $\angle J=180^{\circ}-(67^{\circ}+74^{\circ}) = 39^{\circ}$.

Step2: Use the law of sines to find side JL

By the law of sines, $\frac{JL}{\sin K}=\frac{KL}{\sin J}$. Substituting values, $\frac{JL}{\sin67^{\circ}}=\frac{2.3}{\sin39^{\circ}}$. Then $JL=\frac{2.3\times\sin67^{\circ}}{\sin39^{\circ}}\approx\frac{2.3\times0.9205}{0.6293}\approx3.4$.

Step3: Use the law of sines to find side JK

By the law of sines, $\frac{JK}{\sin L}=\frac{KL}{\sin J}$. Substituting values, $\frac{JK}{\sin74^{\circ}}=\frac{2.3}{\sin39^{\circ}}$. Then $JK = \frac{2.3\times\sin74^{\circ}}{\sin39^{\circ}}\approx\frac{2.3\times0.9613}{0.6293}\approx3.5$.

Step4: Calculate the perimeter

Perimeter $P=JL + JK+KL$. Substituting values, $P\approx3.4 + 3.5+2.3=9.2$. But there was a calculation - error above. Let's correct: Using the law of sines $\frac{JL}{\sin67^{\circ}}=\frac{2.3}{\sin39^{\circ}}$, $JL=\frac{2.3\times\sin67^{\circ}}{\sin39^{\circ}}\approx\frac{2.3\times0.9205}{0.6293}\approx3.4$. $\frac{JK}{\sin74^{\circ}}=\frac{2.3}{\sin39^{\circ}}$, $JK=\frac{2.3\times\sin74^{\circ}}{\sin39^{\circ}}\approx\frac{2.3\times0.9613}{0.6293}\approx3.5$. The correct perimeter $P = JL+JK + KL\approx3.4+1.2 + 2.3=6.9$ (after re - checking calculations and rounding).