Question #5fe15

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Question #5fe15

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Answer 1 · 2017-04-06T01:25:07

$sin x = \frac{4}{5}$ $sinx = 4/5$

Explanation:

Use ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ .

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$

${sin}^{2} x + {(\frac{3}{5})}^{2} = 1$ $sin^2x + (3/5)^2 = 1$

${sin}^{2} x = 1 - \frac{9}{25}$ $sin^2x = 1 - 9/25$

${sin}^{2} x = \frac{16}{25}$ $sin^2x = 16/25$

$sin x = \pm \frac{4}{5}$ $sinx = +- 4/5$

But since the angle is described as being "positive acute", it must be in the firs quadrant, and so the ratio is positive.

Practice Exercises

$1$ $1$ . Knowing that $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ and $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ , find the value of $tan θ$ $tantheta$ if $sec θ = \frac{13}{12}$ $sectheta = 13/12$ and $sin θ > 0$ $sintheta > 0$ .

Solution

$1$ $1$ . $tan θ = \frac{5}{12}$ $tantheta = 5/12$

Hopefully this helps, and good luck!

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Question #5fe15

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