Trigonometry (10th Edition)

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

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 68

Answer

The coordinates of P is $(1,\sqrt 3)$
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Work Step by Step

Draw a perpendicular line from P to x-axis. It cuts the x-axis at Q. O is the origin. OQP is a right triangle, right-angled at Q. $\angle POQ = 60^{\circ} (Given)$ $OP = 2 (Given)$ $\sin\theta = \frac{Opposite Side}{Hypotenuse}$ Using trignometric ratio for $60^{\circ}$ $\sin60^{\circ} = \frac{PQ}{OP}$ Substituting the values for OP and $sin60^{\circ} = \frac{\sqrt 3}{2} $ [Trigonometric Function Value for $sin60^{\circ}$ ] $\frac{\sqrt 3}{2} = \frac{PQ}{2}$ $PQ = \sqrt3$ By Pythagorean Theorem, $OP^{2} = OQ^{2}+PQ^{2}$ $2^{2}= OQ^{2}+(\sqrt 3)^{2}$ $4= OQ^{2}+3$ $ OQ^{2}=4-3$ $ OQ^{2}=1$ $ OQ=1$ The coordinates of P =P(OQ,PQ) = $P(1,\sqrt 3)$
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