Question

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question 1-6
in $\triangle jkl$, $m\angle l = 90^\circ$. if $\sin j = 0.6$, what is the value of $\cos k$ to the nearest tenth?
enter the correct answer in the box.

Explanation:

Step1: Analyze the right triangle

In right triangle \( \triangle JKL \) with \( \angle L = 90^\circ \), the sum of the other two angles \( \angle J \) and \( \angle K \) is \( 90^\circ \) (since the sum of angles in a triangle is \( 180^\circ \), and \( \angle L = 90^\circ \)). So \( \angle J + \angle K = 90^\circ \), which means \( \angle K = 90^\circ - \angle J \).

Step2: Use the co - function identity

We know the co - function identity \( \cos(90^\circ - \theta)=\sin\theta \). Let \( \theta=\angle J \), then \( \cos K=\cos(90^\circ - \angle J) \). By the co - function identity, \( \cos(90^\circ - \angle J)=\sin\angle J \).

Step3: Substitute the given value

We are given that \( \sin J = 0.6 \). Since \( \cos K=\sin J \), then \( \cos K = 0.6 \).

Answer:

\( 0.6 \)