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: 5

Answer

There are two possible values for the angle $A$ $A =54^{\circ}20'$ $A = 125^{\circ}40'$

Work Step by Step

We can use the law of sines to find angle $A$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~A = \frac{a~sin~B}{b}$ $A = arcsin(\frac{a~sin~B}{b})$ $A = arcsin(\frac{(340~m)~sin~39^{\circ}50'}{268~m})$ $A = arcsin(\frac{(340~m)~sin~39.83}{268~m})$ $A = arcsin(0.81259)$ $A = 54.35^{\circ} = 54^{\circ}20'$ Note that another possible value for angle $A$ is $180^{\circ}- 54^{\circ}20'$ which is $125^{\circ}40'$
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