Answer:
[tex]k \approx 5$ units[/tex]
Step-by-step explanation:
In Triangle JKL
[tex]\angle K=120^\circ\\\angle L=40^\circ\\KL=2\\JL=k[/tex]
We want to determine the approximate value of k using the law of sines.
[tex]\angle J+\angle K+\angle L=180^\circ $ (Sum of angles in a \triangle)\\\angle J+120^\circ+40^\circ=180^\circ \\\angle J=180^\circ-(120^\circ+40^\circ)=20^\circ[/tex]
Using Law of Sines
[tex]\dfrac{k}{\sin K} =\dfrac{j}{\sin J} \\\dfrac{k}{\sin 120} =\dfrac{2}{\sin 20} \\k=\sin 120 \times \dfrac{2}{\sin 20}\\k=5.06\\k \approx 5$ units[/tex]
