Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 7 - Applications of Trigonometric Functions - Section 7.4 Area of a Triangle - 7.4 Assess Your Understanding - Page 565: 28

Answer

$1.89$

Work Step by Step

We know that the area of a triangle with sides $a,b$ and $c$ and their opposite angles $A, B$ and $C$ can be represented as: $K=\dfrac{a^2sin(B)\sin(C)}{2\sin(A)}; \\ K=\dfrac{b^2\sin(A)\sin(C)}{2\sin(B)}; \\ K=\dfrac{c^2\sin(A)\sin(B)}{2\sin(C)}$ Remember that the sum of the angles of the triangle is $180^{\circ}$ Therefore, $B=180^{\circ}-50^{\circ}-20^{\circ}=110^{\circ}$ Now, we have: $K=\dfrac{a^2sin(B)\sin(C)}{2\sin(A)}=\dfrac{(3)^2\sin(110^{\circ})\sin(20^{\circ})}{2\sin(50^{\circ})} \approx 1.89$
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