Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises: 14

Answer

$\angle C= 91.9^{\circ}$, $BC \approx 490$ ft, $AB \approx 848$ ft

Work Step by Step

1. Find $\angle C$ $\angle C = 180 - (\angle B + \angle A)$ $= 180 - (52.8 + 35.3)$ $= 180 - 88.1$ $= 91.9^{\circ}$ 2. Find $BC$ $\frac{BC}{sin(A)} = \frac{AC}{sin(B)}$ $\frac{BC}{sin(35.3)} = \frac{675}{sin(52.8)}$ $BC = \frac{675sin(35.3)}{sin(52.8)}$ by GDC / calculator $BC = 489.69...$ $BC \approx 490$ ft 2. Find $AB$ $\frac{AB}{sin(C)} = \frac{AC}{sin(B)}$ $\frac{AB}{sin(91.9)} = \frac{675}{sin(52.8)}$ $AB = \frac{675sin(91.9)}{sin(52.8)}$ by GDC / calculator $= 847.95...$ $AB \approx 848$ ft
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