Given:
IJK is the right triangle.
[tex]\angle J=90^{\circ},\angle I=60^{\circ},IK=2\sqrt[]{21}[/tex]
Use the cosine ratio,
[tex]\begin{gathered} \cos I=\frac{adjacent\text{ side}}{\text{hypotenuse}} \\ \cos 60^{\circ}=\frac{IJ}{IK} \\ \cos 60^{\circ}=\frac{IJ}{2\sqrt[]{21}} \\ \frac{1}{2}=\frac{IJ}{2\sqrt[]{21}} \\ IJ=\frac{2\sqrt[]{21}}{2} \\ IJ=\sqrt[]{21} \end{gathered}[/tex]
Answer:
[tex]IJ=\sqrt[]{21}[/tex]
