Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.4 Area of a Triangle - 8.4 Assess Your Understanding - Page 544: 31


$2.27$ square units

Work Step by Step

We know that $K=\frac{a^2\sin(B)\sin(C)}{2\sin(A)}=\frac{b^2\sin(A)\sin(C)}{2\sin(B)}=\frac{c^2\sin(A)sin(B)}{2\sin(C)}$ and also that the sum of the angles of the triangle is $180^o.$ Hence, $A=180^o-70^o-10^o=100^o.$ Thus, $K=\frac{5^2\sin(100^o)\sin(10^o)}{2\sin(70^o)}\approx 2.27.$
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