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: 40


$a = 0.6006~cm$ $b = 4.787~cm$ $\angle A = 7^{\circ}09'$

Work Step by Step

We can convert angle B to degrees: $B = 82^{\circ}51' = (82+\frac{51}{60})^{\circ} = 82.85^{\circ}$ We can use angle B and angle C to find angle A: $\angle A = 180^{\circ}-90^{\circ}-82.85^{\circ} = 7.15^{\circ}$ $\angle A = 7.15^{\circ}$ which is $7^{\circ}09'$ We can use angle B and $c$ to find $a$: $cos~B = \frac{a}{c}$ $a = (c)~cos~B$ $a = (4.825~cm)~cos(82.85^{\circ})$ $a = 0.6006~cm$ We can use the Pythagorean theorem to find $b$: $b^2 = c^2-a^2$ $b = \sqrt{c^2-a^2}$ $b = \sqrt{(4.825~cm)^2-(0.6006~cm)^2}$ $b = 4.787~cm$
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