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 324: 63


$A = 24\sqrt{3}$ We also find this area when we use Heron's formula.

Work Step by Step

We can find the area of the triangle: $A = \frac{1}{2}bh$ $A = \frac{1}{2}(16)(3\sqrt{3})$ $A = 24\sqrt{3}$ We can find the semiperimeter of the triangle: $S = \frac{a+b+c}{2}$ $S = \frac{6+14+16}{2}$ $S = 18$ We can use Heron's formula to find the area of the triangle: $A = \sqrt{S(S-a)(S-b)(S-c)}$ $A = \sqrt{18(18-6)(18-14)(18-16)}$ $A = \sqrt{1728}$ $A = 24\sqrt{3}$
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