Given the figure in the attached image.
The triangle JKL is a right-angled triangle.
Applying trigonometry;
[tex]\cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}}[/tex]
Given;
[tex]\begin{gathered} \theta=30^{\circ} \\ \text{adjacent = }JK \\ \text{hypotenuse = KL} \end{gathered}[/tex]
Substituting the given values;
[tex]\begin{gathered} \cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}} \\ \cos 30=\frac{JK}{KL} \end{gathered}[/tex]
Cross multiply;
[tex]JK=(\cos 30)KL[/tex]
Recall that;
[tex]\cos 30=\frac{\sqrt[]{3}}{2}[/tex]
So, we have;
[tex]\begin{gathered} JK=(\cos 30)KL \\ JK=\frac{\sqrt[]{3}}{2}KL \end{gathered}[/tex]
Therefore, the equ
