Trigonometry (11th Edition) Clone

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

Chapter 7 - Review Exercises - Page 349: 9b


A has two possible values when $5 \lt b \lt 10$

Work Step by Step

The sum of the three angles in the triangle is $180^{\circ}$. Since $B = 30^{\circ},$ then $0^{\circ} \lt A \lt 150^{\circ}$ We know that $sin~\theta = sin~(180^{\circ}-\theta)$. Therefore, the angle $A$ has two possible values when $30^{\circ} \lt A \lt 90^{\circ}$ or $90^{\circ} \lt A \lt 150^{\circ}$ Then $0.5 \lt sin~A \lt 1$ We can use the law of sines to find $b$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(10)~sin~30^{\circ}}{sin~A}$ $b = \frac{5}{sin~A}$ If $0.5 \lt sin~A \lt 1$ then $5 \lt b \lt 10$ Therefore, A has two possible values when $5 \lt b \lt 10$
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