Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.1 The Law of Sines - 8.1 Exercises - Page 757: 81

Answer

$$65.94{\text{ c}}{{\text{m}}^2}$$

Work Step by Step

$$\eqalign{ & A = {\text{3}}0.{\text{5}}0^\circ ,b = {\text{13}}.00{\text{ cm}},C = {\text{112}}.{\text{6}}0^\circ \cr & {\text{Find the angle }}B \cr & B = 180^\circ - A - C \cr & B = 180^\circ - 30.50^\circ - 112.60^\circ \cr & B = 36.9^\circ \cr & \cr & {\text{Use the law of sines to find side }}a \cr & \frac{a}{{\sin A}} = \frac{b}{{\sin B}} \cr & a = \frac{{b\sin A}}{{\sin B}} \cr & a = \frac{{13\sin \left( {30.50^\circ } \right)}}{{\sin \left( {36.9^\circ } \right)}} \cr & a \approx 10.98841{\text{ cm}} \cr & \cr & {\text{Find the area of the triangle using the formula }}Area = \frac{1}{2}ab\sin C \cr & Area = \frac{1}{2}\left( {0.98841{\text{ cm}}} \right)\left( {{\text{13}}.00{\text{ cm}}} \right)\sin \left( {112.60^\circ } \right) \cr & Area \approx 65.94{\text{ c}}{{\text{m}}^2} \cr} $$
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