In $$ riangle J K L, \mathrm{~m} \angle L=90^{\circ} $$. If $$ \sin J=0.6 $$, what is the value of $$ \cos K $$ to the nearest tenth? Enter the correct answer in the box.

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In $$ 	riangle J K L, \mathrm{~m} \angle L=90^{\circ} $$. If $$ \sin J=0.6 $$, what is the value of $$ \cos K $$ to the nearest tenth? Enter the correct answer in the box.

UpStudy AI · Accepted Answer

In triangle $$ \triangle JKL $$ with $$ \angle L = 90^\circ $$, the angles $$ J $$ and $$ K $$ are complementary, meaning they add up to $$ 90^\circ $$. Therefore, $$ \angle K = 90^\circ - \angle J $$.

Given that $$ \sin J = 0.6 $$, we can use the co-function identity:
$$
\cos K = \cos(90^\circ - J) = \sin J = 0.6
$$

So, the value of $$ \cos K $$ to the nearest tenth is:

**0.6**
$$ \cos K = 0.6 $$

In . If , what is the value of to the nearest tenth?
Enter the correct answer in the box.

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In . If , what is the value of to the nearest tenth? Enter the correct answer in the box.