Question #70b87

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Question #70b87

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Answer 1 · 2017-05-13T16:57:52

If Quadrant I: $\frac{4}{5}$ $4/5$
If Quadrant II: $- \frac{4}{5}$ $-4/5$

Explanation:

Use Pythagorean's Identity: ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

By substitution, we get:
${(\frac{3}{5})}^{2} + {cos}^{2} a = 1$ $(3/5)^2+cos^2a=1$

Subtract ${(\frac{3}{5})}^{2}$ $(3/5)^2$ from both sides:
${cos}^{2} a = 1 - {(\frac{3}{5})}^{2}$ $cos^2a=1-(3/5)^2$
${cos}^{2} a = 1 - \frac{9}{25}$ $cos^2a=1-9/25$
${cos}^{2} a = \frac{16}{25}$ $cos^2a=16/25$

Now, we take the square root of both sides:
$cos a = \pm \sqrt{\frac{16}{25}}$ $cosa=+-sqrt(16/25)$
$cos a = \pm \frac{4}{5}$ $cosa=+-4/5$

Therefore, if $∠ A$ $angleA$ is in Quadrant I, $cos a = \frac{4}{5}$ $cosa=4/5$
if $∠ A$ $angleA$ is in Quadrant II, $cos a = - \frac{4}{5}$ $cosa=-4/5$

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Question #70b87

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