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 7.3 The Law of Cosines - 7.3 Exercises - Page 323: 60

Answer

The distance between the airplane and the mountain is 5.99 km when the second bearing is taken.

Work Step by Step

Let A be the point where the mountain is at a bearing of $24.1^{\circ}$ Let B be the point where the mountain is at a bearing of $32.7^{\circ}$ Let point C be the location of the mountain. The points ABC form a triangle. The angle $A = 24.1^{\circ}$ The angle $B = 180^{\circ}-32.7^{\circ} = 147.3^{\circ}$ We can use the law of sines to find the distance $BC$, which is the distance from the mountain when the second bearing is taken: $\frac{BC}{sin~24.1^{\circ}} = \frac{7.92}{sin~147.3^{\circ}}$ $BC = \frac{7.92~sin~24.1^{\circ}}{sin~147.3^{\circ}}$ $BC = 5.99~km$ The distance between the airplane and the mountain is 5.99 km when the second bearing is taken.
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.