## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.1 Oblique Triangles and the Law of Sines - 7.1 Exercises - Page 303: 42

#### Answer

The distance between the centers of atoms A and C is 11.6

#### Work Step by Step

We can find angle $C$: $\frac{a}{sin~A} = \frac{c}{sin~C}$ $sin~C = \frac{c~sin~A}{a}$ $C = arcsin(\frac{c~sin~A}{a})$ $C = arcsin(\frac{(3.0+4.5)~sin~(18^{\circ})}{2.0+3.0})$ $C = arcsin(\frac{(7.5)~sin~(18^{\circ})}{5.0})$ $C = 27.6^{\circ}$ We can find angle $B$: $A+B+C = 180^{\circ}$ $B = 180^{\circ}-A-C$ $B = 180^{\circ}-18^{\circ}-27.6^{\circ}$ $B = 134.4^{\circ}$ We can find the length of side $b$: $\frac{b}{sin~B} = \frac{a}{sin~A}$ $b = \frac{a~sin~B}{sin~A}$ $b = \frac{(5.0)~sin~(134.4^{\circ})}{sin~(18^{\circ})}$ $b = 11.6$ The distance between the centers of atoms A and C is 11.6

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