Question - Solving for the Sine of an Angle in Similar Triangles

Similar triangles have corresponding angles that are equal and corresponding sides that are proportional. Since triangles \( FGH \) and \( JKL \) are similar, and angles \( G \) and \( K \) are right angles, then the triangles are right triangles, and we can use trigonometric ratios to solve for our missing sides or angles.

In a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. In \( \triangle JKL \), \( \sin(J) \) is already given as \( \frac{308}{317} \). This means that if \( J \) is the angle opposite side \( KL \) and \( JK \) is the hypotenuse, the ratio is \( \frac{KL}{JK} = \frac{308}{317} \).

Since \( \angle F \) corresponds to \( \angle J \) and the triangles are similar, we know \( \sin(F) = \sin(J) \) because corresponding angles in similar triangles are equal. Therefore:

Given that the triangles are similar, the ratio \( \frac{FG}{FH} \) in \( \triangle FGH \) would be equal to the ratio \( \frac{KL}{JK} \) in \( \triangle JKL \), which is \( \frac{308}{317} \).

Since we are looking for \( \sin(J) \) which is equal to \( \sin(F) \), and we are already given \( \sin(J) \), we do not need to calculate anything further. The answer is simply the value provided:

Hence, the correct answer is B) \( \frac{308}{317} \).