Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Review Exercises - Page 96: 60

Answer

Refer to the triangle below.
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Work Step by Step

Recall: For an angle $\theta$ whose terminal side contains the point $(x, y)$, we have: $\sin{\theta}=\dfrac{y}{r}$ where $r^2=x^2+y^2$ Thus, with $\sin{\theta}=\frac{3}{4}$, we can assume that $y=3$ and $r=4$. Solve for $x$ using the formula above to obtain: \begin{align*} r^2&=x^2+y^2\\ 4^2&=x^2+3^2\\ 4^2-3^2&=x^2\\ 7&=x^2\\ \pm \sqrt{7}&=x \end{align*} Hence, one triangle whose solution is found from $\sin{\theta}=\dfrac{3}{4}$ is one where $x=\sqrt7$ and $y=3$. Refer to the triangle above.
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