Trigonometry (11th Edition) Clone

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 303: 41

Answer

$\theta = 111.4^{\circ}$

Work Step by Step

Let angle $A = 38^{\circ}$. Then the side $a$ opposite angle $A$ has a length of $a = 3.6+1.6 = 5.2$ Let angle $B$ be the angle at the center of the gear with radius 3.6. Then the side $b$ opposite angle $B$ has a length of $b = 2.7+1.6 = 4.3$ We can find angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $B = arcsin(\frac{b~sin~A}{a})$ $B = arcsin(\frac{(4.3)~sin~(38^{\circ})}{5.2})$ $B = 30.6^{\circ}$ We can find angle $\theta$: $A+B+\theta = 180^{\circ}$ $\theta = 180^{\circ}-A-B$ $\theta = 180^{\circ}-38^{\circ}-30.6^{\circ}$ $\theta = 111.4^{\circ}$
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