Trigonometry (10th Edition)

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

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises: 61

Answer

$P$ = (-3$\sqrt3$,3)
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Work Step by Step

The Point $P$ is represented by the coordinate ($x$,$y$). $\cos$ $\theta$ = $\frac{x}{r}$ $\cos$ 150$^{\circ}$ = $\frac{x}{6}$ x = 6 $\cos$ 150$^{\circ}$ x = 6(-$\frac{\sqrt3}{2}$) x = -3$\sqrt3$ $\sin$ $\theta$ = $\frac{y}{r}$ $\sin$ 150$^{\circ}$ = $\frac{y}{6}$ y = 6 $\sin$ 150$^{\circ}$ y = 6($\frac{1}{2}$) y = 3 Therefore: $P$ = (-3$\sqrt3$,3)
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