Trigonometry (11th Edition) Clone

Published by Pearson

Chapter 7 - Applications of Trigonometry and Vectors - Summary Exercises on Applications of Trigonometry and Vectors - Page 345: 10

Answer

There are two possible triangles. The angles of one triangle are as follows: $A = 28^{\circ}10', B = 36^{\circ}10',$ and $C = 115^{\circ}40'$ The lengths of the sides are as follows: $a = 21.2~yd, b = 26.5~yd,$ and $c = 40.5~yd$ The angles of the other possible triangle are as follows: $A = 28^{\circ}10', B = 143^{\circ}50',$ and $C = 8^{\circ}$ The lengths of the sides are as follows: $a = 21.2~yd, b = 26.5~yd,$ and $c = 6.3~yd$

Work Step by Step

Let $a = 21.2~yd$ Let $b = 26.5~yd$ Let angle $A = 28^{\circ}10'$ Let angle $B$ be the angle that subtends the side $b$. We can use the law of sines to find $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(26.5)~sin~28^{\circ}10'}{21.2}$ $sin~B = 0.59$ $B = arcsin(0.59)$ $B = 36^{\circ}10'$ We can find angle C: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-28^{\circ}10'-36^{\circ}10'$ $C = 115^{\circ}40'$ We can use the law of sines to find $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(21.2~yd)~sin~115^{\circ}40'}{sin~28^{\circ}10'}$ $c = 40.5~yd$ Note that there is another possible value for the angle $B$: $B = 180^{\circ}- 36^{\circ}10' = 143^{\circ}50'$ We can find angle C: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-28^{\circ}10'-143^{\circ}50'$ $C = 8^{\circ}$ We can use the law of sines to find $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(21.2~yd)~sin~8^{\circ}}{sin~28^{\circ}10'}$ $c = 6.3~yd$

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