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 2.4 Solving Right Triangles - 2.4 Exercises - Page 78: 39


$a = 0.4832~m$ $b = 0.3934~m$ $A = 50^{\circ}51'$

Work Step by Step

We can convert angle B to degrees: $B = 39^{\circ}09' = (39+\frac{9}{60})^{\circ} = 39.15^{\circ}$ We can use angle B and angle C to find angle A: $A = 180^{\circ}-90^{\circ}-39.15^{\circ} = 50.85^{\circ}$ $A = 50.85^{\circ}$ which is $50^{\circ}51'$ We can use angle B and $c$ to find $a$: $cos~B = \frac{a}{c}$ $a = (c)~cos~B$ $a = (0.6231~m)~cos(39.15^{\circ})$ $a = 0.4832~m$ We can use the Pythagorean theorem to find $b$: $b^2 = c^2-a^2$ $b = \sqrt{c^2-a^2}$ $b = \sqrt{(0.6231~m)^2-(0.4832~m)^2}$ $b = 0.3934~m$
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