Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Test: 8

Answer

There are two possible solutions for this triangle. The angles of the triangle are as follows: $A = 83^{\circ}, B = 58^{\circ}30',$ and $C = 38^{\circ}30'$ The lengths of the sides are as follows: $a = 1252~in, b = 1075~in,$ and $c = 785~in$ There is another possible solution for this triangle: The angles of the triangle are as follows: $A = 20^{\circ}, B = 121^{\circ}30',$ and $C = 38^{\circ}30'$ The lengths of the sides are as follows: $a = 431~in, b = 1075~in,$ and $c = 785~in$

Work Step by Step

$C = 38^{\circ}30' = 38.5^{\circ}$ $b = 1075~in$ $c = 785~in$ We can use the law of sines to find angle $B$: $\frac{c}{sin~C} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~C}{c}$ $B = arcsin(\frac{b~sin~C}{c})$ $B = arcsin(\frac{1075~sin~38.5^{\circ}}{785})$ $B = arcsin(0.8525)$ $B = 58.5^{\circ} = 58^{\circ}30'$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-58^{\circ}30'-38^{\circ}30'$ $A = 83^{\circ}$ We can use the law of sines to find the length of side $a$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $a = \frac{c~sin~A}{sin~C}$ $a = \frac{(785~in)~sin~(83^{\circ})}{sin~38.5^{\circ}}$ $a = 1252~in$ Note that there is another possibility for angle $B$: $B = 180^{\circ}- 58.5^{\circ} = 121.5^{\circ} = 121^{\circ}30'$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-121^{\circ}30'-38^{\circ}30'$ $A = 20^{\circ}$ We can use the law of sines to find the length of side $a$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $a = \frac{c~sin~A}{sin~C}$ $a = \frac{(785~in)~sin~(20^{\circ})}{sin~38.5^{\circ}}$ $a = 431~in$
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