Use the diagram to complete the statement. L k 52^circ 3B^ast Given Delta JKL,sin(38^circ ) equals cos(38^circ ) cos(52^circ ) tan(38^circ ) tan(52^circ )

In a right triangle, the sine of an angle is equal to the cosine of its complementary angle. Complementary angles are two angles that add up to \(90^\circ\).<br /><br />Given that \(\sin(38^\circ)\) is being compared, we need to find the cosine of the complementary angle to \(38^\circ\). The complementary angle is:<br /><br />\[ 90^\circ - 38^\circ = 52^\circ \]<br /><br />Therefore, \(\sin(38^\circ) = \cos(52^\circ)\).<br /><br />So, the correct answer is:<br /><br />\[ \cos(52^\circ) \]