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 303: 38


The distance across the river is 293.4 meters

Work Step by Step

Let $A$ be Mark's position. Then angle $A = 180^{\circ} - 115.45^{\circ}$ which is $64.55^{\circ}$. Let $C$ be Lisa's position. Then angle $C = 45.47^{\circ}$. Let the tree be located at the position of angle $B$. We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-64.55^{\circ}-45.47^{\circ}$ $B = 69.98^{\circ}$ We can find the length of side $a$ which is the distance from Lisa to the tree: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $a = \frac{b~sin~A}{sin~B}$ $a = \frac{(428.3~m)~sin~(64.55^{\circ})}{sin~(69.98^{\circ})}$ $a = 411.6~mi$ Let angle $\theta$ be the angle between a horizontal line across the river and the line from Lisa to the tree. We can find $\theta$: $\theta = 90^{\circ} - 45.47^{\circ} = 44.53^{\circ}$ We can find the distance $d$ across the river: $\frac{d}{a} = cos~\theta$ $d = a~cos~\theta$ $d = (411.6~m)~cos~(44.53^{\circ})$ $d = 293.4~m$ The distance across the river is 293.4 meters
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