Trigonometry (11th Edition) Clone

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

Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 78: 33


$A = 62.8^{\circ}$ which is $62^{\circ}48'$ $B = 27.2^{\circ}$ which is $27^{\circ}12'$ $c = 85.9 ~yd$

Work Step by Step

We can use the Pythagorean theorem to find c: $c^2 = a^2+b^2$ $c = \sqrt{a^2+b^2}$ $c = \sqrt{(76.4~yd)^2+(39.3~yd)^2}$ $c = 85.9~yd$ We can use $a$ and $b$ to find angle A: $tan~A = \frac{a}{b}$ $tan~A = \frac{76.4}{39.3}$ $A = tan^{-1}(\frac{76.4}{39.3})$ $A = 62.8^{\circ}$ which is $62^{\circ}48'$ We can use angle A and angle C to find angle B: $B = 180^{\circ}-90^{\circ}-62.8^{\circ} = 27.2^{\circ}$ $B = 27.2^{\circ}$ which is $27^{\circ}12'$
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.