Answer
no triangle.
Work Step by Step
1. Use the Law of Sines $\frac{3}{sin40^\circ}=\frac{7}{sinB} $
2. Thus $B_1=sin^{-1}(\frac{7sin40^\circ}{3})\approx sin^{-1}(1.50)$ no solution, thus no triangle.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 7 - Applications of Trigonometric Functions - Chapter Test - Page 580: 7
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