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 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 311: 38


The heading of $Z$ from $X$ is $~~75^{\circ}$ The distance $YZ$ is $~~273.6~miles$

Work Step by Step

The points $X$, $Y,$ and $Z$ form a triangle. We can find $\angle XYZ$: $\angle XYZ = 35^{\circ}+(180^{\circ}-145^{\circ}) = 70^{\circ}$ Since $XY = XZ = 400~mi$, the triangle is an isosceles triangle. Then $\angle XZY = \angle XYZ = 70^{\circ}$ We can find $\angle YXZ$: $\angle YXZ = 180^{\circ} - \angle XYZ - \angle XZY$ $\angle YXZ = 180^{\circ} - 70^{\circ} - 70^{\circ}$ $\angle YXZ = 40^{\circ}$ We can find the heading of $Z$ from $X$: $35^{\circ}+40^{\circ} = 75^{\circ}$ The heading of $Z$ from $X$ is $~~75^{\circ}$ We can find the distance $YZ$: $\frac{YZ}{sin~\angle YXZ} = \frac{XY}{sin~\angle XZY}$ $YZ = \frac{XY~sin~\angle YXZ}{sin~\angle XZY}$ $YZ = \frac{400~sin~40^{\circ}}{sin~70^{\circ}}$ $YZ = 273.6~mi$ The distance $YZ$ is $~~273.6~miles$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.