Given triangle JKL, sin 38 degree equals

Answer:
B. [tex]\text{cos}(52^{\circ})[/tex].
Step-by-step explanation:
We have been given a right triangle. We are asked to complete our given statement.
[tex]\text{sin}(38^{\circ})[/tex] equals ___.
We will use identity [tex]\text{sin}(x)=\text{cos}(90-x)[/tex] to solve our given problem.
[tex]\text{sin}(38^{\circ})=\text{cos}(90^{\circ}-38^{\circ})[/tex]
[tex]\text{sin}(38^{\circ})=\text{cos}(52^{\circ})[/tex]
Therefore, [tex]\text{sin}(38^{\circ})[/tex] equals [tex]\text{cos}(52^{\circ})[/tex].