1. **Problem statement:** Find the exact value of $\cos L$ in simplest radical form for a right triangle $\triangle LKJ$ with right angle at $K$, sides $LK=2$, $KJ=6$, and hypotenuse $LJ=\sqrt{40}$.\n\n2. **Recall the definition of cosine:** For angle $L$, $\cos L = \frac{\text{adjacent side to } L}{\text{hypotenuse}}$.\n\n3. **Identify the sides relative to angle $L$:**\n- Adjacent side to $L$ is $LK=2$.\n- Hypotenuse is $LJ=\sqrt{40}$.\n\n4. **Write the cosine ratio:**\n$$\cos L = \frac{2}{\sqrt{40}}$$\n\n5. **Rationalize the denominator:**\nMultiply numerator and denominator by $\sqrt{40}$:\n$$\cos L = \frac{2}{\sqrt{40}} \times \frac{\sqrt{40}}{\sqrt{40}} = \frac{2\sqrt{40}}{40}$$\n\n6. **Simplify $\sqrt{40}$:**\n$$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$$\n\n7. **Substitute back and simplify:**\n$$\cos L = \frac{2 \times 2 \sqrt{10}}{40} = \frac{4 \sqrt{10}}{40}$$\n\n8. **Reduce the fraction:**\n$$\cos L = \frac{\cancel{4} \sqrt{10}}{\cancel{40}} = \frac{\sqrt{10}}{10}$$\n\n**Final answer:**\n$$\boxed{\cos L = \frac{\sqrt{10}}{10}}$$