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 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 304: 48


$\sqrt 3$ sq. units

Work Step by Step

We first use the formula $A=\frac{1}{2}bh$ to find the area of the triangle: $A=\frac{1}{2}bh$ $A=\frac{1}{2}(2)(\sqrt 3)$ $A=\sqrt 3$ Now, we use the formula $Area=\frac{1}{2}ab \sin C$ to verify the result: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(2)(2) \sin 60$ $Area=\frac{4}{2} \sin 60$ $Area=2(\sin60)$ $Area=2(\frac{\sqrt 3}{2})$ $Area=\sqrt 3$ Therefore, both the formulas give the same area which is $\sqrt 3$ sq. units.
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