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: 71

Answer

$$\frac{{\sqrt 3 }}{2}{\text{ sq unit}}$$

Work Step by Step

$$\eqalign{ & {\text{From the triangle we have:}} \cr & A = {60^ \circ },\,\,\,C = {90^ \circ } \cr & a = \sqrt 3 ,\,\,\,b = 1,\,\,\,c = 2,\,\,\,h = \sqrt 3 \cr & {\text{Find the area of the triangle using the formula }}Area = \frac{1}{2}bh \cr & Area = \frac{1}{2}\left( 1 \right)\left( {\sqrt 3 } \right) \cr & Area = \frac{{\sqrt 3 }}{2}{\text{ sq unit}} \cr & {\text{Find the area of the triangle using the formula }}Area = \frac{1}{2}ab\sin C \cr & Area = \frac{1}{2}\left( {\sqrt 3 } \right)\left( 1 \right)\sin \left( {{{90}^ \circ }} \right) \cr & Area = \frac{{\sqrt 3 }}{2}{\text{ sq unit}} \cr & {\text{Both results are the same}} \cr} $$
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