Trigonometry (10th Edition)

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

Chapter 7 - Applications of Trigonometry and Vectors - Summary Exercises on Applications of Trigonometry and Vectors - Page 338: 2

Answer

The distance between the wrist pin and the connecting center is 38.3 cm

Work Step by Step

Let $w = 11.2~cm$ Let $c = 28.6~cm$ The three points P, C, and W form a triangle. We can use the law of sines to find the angle W: $\frac{c}{sin~C} = \frac{w}{sin~W}$ $sin~W = \frac{w~sin~C}{c}$ $W = arcsin(\frac{w~sin~C}{c})$ $W = arcsin(\frac{(11.2)~sin~25.5^{\circ}}{28.6})$ $W = arcsin(0.16859)$ $W = 9.7^{\circ}$ We can find the angle P: $C+W+P = 180^{\circ}$ $P = 180^{\circ} - C - W$ $P = 180^{\circ} - 25.5^{\circ} - 9.7^{\circ}$ $P = 144.8^{\circ}$ We can use the law of sines to find p, which is the distance between the wrist pin and the connecting center: $\frac{p}{sin~P} = \frac{c}{sin~C}$ $p = \frac{c~sin~P}{sin~C}$ $p = \frac{(28.6~cm)~sin~144.8^{\circ}}{sin~25.5^{\circ}}$ $p = 38.3~cm$ The distance between the wrist pin and the connecting center is 38.3 cm
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