- Pythagorean Theorem and Special Right Triangles, Trigonometry for Right TrianglesChoice D
$$ JL=\sqrt{9^2+7^2}=\sqrt{130} $$
$$ \therefore cos\ KJL=\frac{KJ}{JL}=\frac{9}{\sqrt{130}} $$
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$$ JL=\sqrt{9^2+7^2}=\sqrt{130} $$
$$ \therefore cos\ KJL=\frac{KJ}{JL}=\frac{9}{\sqrt{130}} $$
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