University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 62


$$\sin B=\frac{3\sqrt{21}}{14}$$

Work Step by Step

$$a = 2 \hspace{1cm}b=3\hspace{1cm}C=60^\circ$$ - From Exercise 61, the law of sines: $$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$ To find $\sin B$, we still need to find $c$. And we can use the law of cosines to do that: $$c^2=a^2+b^2-2ab\cos C$$ Therefore, $$c^2=2^2+3^2-2\times2\times3\times\cos60^\circ$$ $$c^2=4+9-12\times\frac{1}{2}$$ $$c^2=13-6$$ $$c^2=7$$ $$c=\sqrt7$$ (as $c\gt0$) - Now to find $\sin B$, we use the law of sines: $$\frac{\sin B}{b}=\frac{\sin C}{c}$$ $$\sin B=\frac{b\times\sin C}{c}$$ $$\sin B=\frac{3\times\sin60^\circ}{\sqrt7}$$ $$\sin B=\frac{3\times\frac{\sqrt3}{2}}{\sqrt7}$$ $$\sin B=\frac{3\sqrt3}{2\sqrt7}$$ $$\sin B=\frac{3\sqrt{21}}{14}$$
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