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given right triangle jkl, what is the value of cos(l)? options: \\(\\fr…

Question

given right triangle jkl, what is the value of cos(l)? options: \\(\frac{5}{13}\\), \\(\frac{5}{12}\\), \\(\frac{12}{13}\\), \\(\frac{12}{5}\\) (triangle jkl with right angle at k, jk=12, kl=5)

Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$.

Step2: Find hypotenuse (JL)

Using Pythagorean theorem: $JL = \sqrt{JK^2 + KL^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.

Step3: Identify sides for $\cos(L)$

For $\angle L$, adjacent side is $KL = 5$, hypotenuse is $JL = 13$. So $\cos(L) = \frac{5}{13}$.

Answer:

$\frac{5}{13}$ (corresponding to the first option)