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: 63

Answer

$\tan$ 60$^{\circ}$ = $\sqrt 3$

Work Step by Step

$\tan$ 60$^{\circ}$ We must find side $x$ of the 30$^{\circ}$ - 60$^{\circ}$ Right Triangle. Pythagorean Theorem: $c$$^{2}$ = $a$$^{2}$ + $b$$^{2}$ 2$^{2}$ = 1$^{2}$ + $x$$^{2}$ 4 = 1 + $x$$^{2}$ x$^{2}$ = 3 x = $\sqrt 3$ Now consider the triangle from the perspective of the 60$^{\circ}$ angle. Hypotenuse = 2 Opposite = $\sqrt 3$ Adjacent = 1 $\tan$ 60$^{\circ}$ = $\frac{Opposite}{Adjacent}$ = $\frac{\sqrt 3}{1}$ = $\sqrt 3$ Therefore: $\tan$ 60$^{\circ}$ = $\sqrt 3$
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