Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 306: 41

Answer

$A = \left[ 5~\frac{sin~A~sin~B}{sin~(A+B)}\right]~R^2$

Work Step by Step

The angle $C = 180^{\circ}-(A+B)$ $sin~C = sin~(180^{\circ}-(A+B))$ $sin~C = sin~(180^{\circ})~cos~(A+B)-sin~(A+B)~cos~(180^{\circ})$ $sin~C = (0)~cos~(A+B)-sin~(A+B)~(-1)$ $sin~C = sin~(A+B)$ We can use the law of sines to write an expression for $r$: $\frac{r}{sin~A} = \frac{R}{sin~C}$ $r = \frac{R~sin~A}{sin~C}$ $r = \frac{R~sin~A}{sin~(A+B)}$ Let $x$ be the length of the side AC. We can use the law of sines to write an expression for $x$: $\frac{x}{sin~B} = \frac{R}{sin~C}$ $x = \frac{R~sin~B}{sin~C}$ $x = \frac{R~sin~B}{sin~(A+B)}$ We can find the area of the triangle ABC: $Area = \frac{1}{2}r~x~sin~C$ $Area = \frac{1}{2}~(\frac{R~sin~A}{sin~(A+B)})~(\frac{R~sin~B}{sin~(A+B)})~sin~(A+B)$ $Area = \frac{1}{2}~\frac{R^2~sin~A~sin~B}{sin~(A+B)}$ $Area = \frac{1}{2}~\frac{sin~A~sin~B}{sin~(A+B)}~R^2$ We can find the total area $A$ of ten such triangles: $A = 10 \times \frac{1}{2}~\frac{sin~A~sin~B}{sin~(A+B)}~R^2$ $A = \left[ 5~\frac{sin~A~sin~B}{sin~(A+B)}\right]~R^2$
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