Trigonometry (10th Edition)

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

Chapter 7 - Test - Page 348: 16

Answer

The distance between the transmitter and point A is 1.91 miles

Work Step by Step

Let point C be the position of the transmitter. The points ABC form a triangle. We can find the angle at point A: $A = 90^{\circ}-48^{\circ} = 42^{\circ}$ We can find the angle at point B: $B = 302^{\circ}-270^{\circ} = 32^{\circ}$ We can find the angle at point C: $C = 180^{\circ}-42^{\circ} -32^{\circ} = 106^{\circ}$ We can use the law of sines to find the distance between the transmitter and point A: $\frac{AC}{sin~B} = \frac{AB}{sin~C}$ $AC = \frac{AB~sin~B}{sin~C}$ $AC = \frac{(3.46~mi)~sin~32^{\circ}}{sin~106^{\circ}}$ $AC = 1.91~mi$ The distance between the transmitter and point A is 1.91 miles
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