Trigonometry (11th Edition) Clone

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 302: 34

Answer

$$RS\approx442.6yd$$

Work Step by Step

$$\angle R=102^\circ20'\hspace{.75cm}\angle T=32^\circ50'\hspace{.7cm}TR=582yd$$ 1) Analysis - To find $RS$, we would use law of sines. 2 groups of side and its opposite angles must be known. - Side $TR$ is known, but its opposite angle $\angle S$ is unknown now. However, since $\angle T$ and $\angle R$ are known, $\angle S$ can be calculated using the law of the sum of 3 angles in a triangle. - The opposite angle of $RS$, $\angle T$, is already known. 2) Calculate $\angle S$ The sum of 3 angles in a triangle is $180^\circ$. $$\angle R+\angle T+\angle S=180^\circ$$ $$\angle S+102^\circ20'+32^\circ50'=180^\circ$$ $$\angle S+135^\circ10'=180^\circ$$ $$\angle S=44^\circ50'$$ 3) Now we apply law of sines to find $RS$. - The opposite angle of $RS$ is $\angle T$, $\sin T=\sin 32^\circ50'\approx0.54$ - $TR=582yd$, its opposite angle is $\angle S$, $\sin S=\sin 44^\circ50'\approx0.71$ According to law of sines: $$\frac{RS}{\sin T}=\frac{TR}{\sin S}$$ $$RS=\frac{TR\sin T}{\sin S}$$ $$RS=\frac{582yd\times0.54}{0.71}$$ $$RS\approx442.6yd$$
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