Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 83

Answer

$c^{2}=a^{2}+b^{2}-2ab cos \theta$

Work Step by Step

Here $x=bcos\theta$ and $y=bsin\theta$ Thus $P(x,y)=P(bcos\theta,bsin\theta)$ Use distance formula to compute $c$. Therefore, $c=\sqrt {(a-bcos\theta)^{2}+(0-bsin \theta)^{2}}$ $=\sqrt {(a^{2}+b^{2}cos^{2}\theta-2ab cos \theta+b^{2}sin ^{2}\theta}$ $=\sqrt {(a^{2}-2ab cos \theta+b^{2}(1)}$ Hence, $c^{2}=a^{2}+b^{2}-2ab cos \theta$
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