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We are asked to find \( \cos L \) in simplest terms for the given right triangle \( \triangle LJK \).

The given triangle is right-angled at \( K \), meaning we can apply trigonometric ratios for angle \( L \). The cosine of an angle in a right triangle is defined as:

\[
\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}
\]

In this case:
- The adjacent side to \( \angle L \) is \( \overline{KJ} = \sqrt{75} \).
- The hypotenuse of the triangle is \( \overline{LJ} = \sqrt{46} \).

Thus, the cosine of angle \( L \) is:

\[
\cos L = \frac{\overline{KJ}}{\overline{LJ}} = \frac{\sqrt{75}}{\sqrt{46}}
\]

Next, simplify the fraction:

\[
\cos L = \frac{\sqrt{75}}{\sqrt{46}} = \frac{\sqrt{25 \times 3}}{\sqrt{46}} = \frac{5\sqrt{3}}{\sqrt{46}}
\]

Since this is in its simplest form, \( \cos L = \frac{5\sqrt{3}}{\sqrt{46}} \) is the required answer.

Would you like further details on this? Here are some related questions you might find helpful:

1. How can we rationalize the denominator in trigonometric expressions?
2. What is the significance of using the cosine ratio in right triangles?
3. Can the Pythagorean theorem help verify the side lengths in right triangles?
4. How would you calculate the sine and tangent of \( \angle L \)?
5. How can trigonometric functions be used in solving real-life problems involving triangles?

**Tip:** Always double-check the right-angle location when using trigonometric ratios in geometry problems!

Express cos L as a fraction in simplest terms for the given right triangle.

Learn how to express cos L as a fraction in simplest terms for a right triangle with side lengths involving square roots. This problem demonstrates the use of the cosine trigonometric ratio, simplified square roots, and the Pythagorean theorem. Suitable for students in Grades 9-11.

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