Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 11 - Section 11.4 - Applications with Acute Triangles - Exercises - Page 528: 31

Answer

The distance from the aircraft to the enemy headquarters is $\approx 8812$ m.

Work Step by Step

1. Find the angles in the left $\triangle$ $\angle C = 90 - 53$ $= 37^{\circ}$ $\angle A = 180 - (37 + 90)$ $= 180 - 127$ $= 53^{\circ}$ 2. Find the angles in the right $\triangle$ $\angle C = 90 - 62$ $= 28^{\circ}$ $\angle B = 180 - (28 + 90)$ $= 180 - 118$ $= 62^{\circ}$ 3. Add the $\angle C$ from both triangles to find $\angle C$ of the combined triangle $\triangle ABC$ $= 37 + 28$ $= 65^{\circ}$ 4. Use the Law of Sines to find the distance from $C$ to $B$ Let $x = \overline{CB}$ $\frac{x}{sin(53)} = \frac{10,000}{sin(65)}$ $x = \frac{10,000sin(53)}{sin(65)}$ by GDC / calculator $x = 8811.96...$ m $x \approx 8812$ m
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