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: 49


$\frac{\sqrt 2}{2}$ sq. units

Work Step by Step

We first use the formula $A=\frac{1}{2}bh$ to find the area of the triangle: $A=\frac{1}{2}bh$ $A=\frac{1}{2}(1)(\sqrt 2)$ $A=\frac{\sqrt 2}{2}$ Now, we use the formula $Area=\frac{1}{2}ab \sin C$ to verify the result: $Area=\frac{1}{2}ab \sin C$ $Area=\frac{1}{2}(1)(2) \sin 45$ $Area=\frac{2}{2} \sin 45$ $Area=1(\sin45)$ $Area=1(\frac{\sqrt 2}{2})$ $Area=\frac{\sqrt 2}{2}$ Therefore, both the formulas give the same area which is $\frac{\sqrt 2}{2}$ sq. units.
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