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

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} \\)

Explanation:

Step1: Recall cosine definition

For an acute angle in a right triangle, $\cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}}$

Step2: Identify sides for $\angle L$

Adjacent to $\angle L$: $KL = 5$; First calculate hypotenuse $JL$ using Pythagorean theorem:
$$JL = \sqrt{JK^2 + KL^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$

Step3: Compute $\cos(L)$

Substitute values into cosine formula:
$$\cos(L) = \frac{\text{Adjacent to } L}{\text{Hypotenuse}} = \frac{5}{13}$$

Answer:

A. $\frac{5}{13}$