Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 13

Answer

$\sin$ 60^{\circ} = $\frac{\sqrt3}{2}$ $\cot$ 60^{\circ} = $\frac{\sqrt3}{3}$ $\csc$ 60^{\circ} = $\frac{2\sqrt3}{3}$

Work Step by Step

60^{\circ} $\sin$ 60^{\circ} = $\frac{Opposite}{Hypotenuse}$ = $\frac{\sqrt3}{2}$ $\cot$ 60^{\circ} = $\frac{Adjacent}{Opposite}$ = $\frac{\sqrt3}{3}$ $\csc$ 60^{\circ} = $\frac{Hypotenuse}{Opposite}$ = $\frac{2\sqrt3}{3}$
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