## Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage

# 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

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.