Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 325: 79

Answer

$A = 9.49$

Work Step by Step

From Exercise 77: $a = \sqrt{34}$ $b = \sqrt{29}$ $c = \sqrt{13}$ We can find the semiperimeter of the triangle: $S = \frac{a+b+c}{2}$ $S = \frac{\sqrt{34}+\sqrt{29}+\sqrt{13}}{2}$ $S = 7.41$ We can use Heron's formula to find the area of the triangle: $A = \sqrt{S(S-a)(S-b)(S-c)}$ $A = \sqrt{7.41(7.41-\sqrt{34})(7.41-\sqrt{29})(7.41-\sqrt{13})}$ $A = \sqrt{90.135}$ $A = 9.49$
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