Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.2 - The Rectangular Coordinate System - 1.2 Problem Set - Page 25: 55


$(-0.8660,0.5), (-0.8660,-0.5)$

Work Step by Step

The line $x=-\frac{\sqrt{3}}{2}$ intersects the circle $x^2+y^2=1$ at $(-0.8660,0.5)\,$and$\,(-0.8660,-0.5)$ as can be seen from the figure. The solution can also be obtained analytically by substituting $x=-\frac{\sqrt{3}}{2}$ in the equation $x^2+y^2=1$ and solving for $y$ $$y^2 = 1-x^2$$ $$y = \pm \sqrt{1-x^2} = \pm \sqrt{1-(-\frac{\sqrt{3}}{2})^2} = \pm \sqrt{0.25}$$ $$\therefore y = \pm \frac{1}{2} = \pm 0.5$$ $\therefore $ The coordinates are $(-0.8660,0.5)\,$and$\,(-0.8660,-0.5)$.
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