Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Review Exercises - Page 96: 61a


$h = 716~mi$

Work Step by Step

$h = R~(\frac{1}{cos~\frac{180~T}{P}}-1)$ We can find $h$ when $T = 25~min$ and $P = 140~min$: $h = R~(\frac{1}{cos~\frac{180~T}{P}}-1)$ $h = (3955~mi)~(\frac{1}{cos~\frac{(180)(25~min)}{140~min}}-1)$ $h = 716~mi$
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.