## Geometry: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 8 - Right Triangles and Trigonometry - 8-6 Law of Cosines - Practice and Problem-Solving Exercises - Page 530: 13

#### Answer

$x \approx 54.1$ $m \angle y \approx 72.0^{\circ}$

#### Work Step by Step

First, we want to know what angle is opposite the side in question. The angle that is opposite to the side we are looking for, $x$, is the angle that measures $40^{\circ}$, so let's plug in what we know into the formula for the law of cosines: $x^2 = 78^2 + 80^2 - 2(78)(80)$ cos $40^{\circ}$ Evaluate exponents first, according to order of operations: $x^2 = 6084 + 6400 - 2(78)(80)$ cos $40^{\circ}$ Add to simplify on the right side of the equation: $x^2 = 12484 - 2(78)(80)$ cos $40^{\circ}$ Take the square root of both sides of the equation: $x \approx 54.1$ Now, we use the law of cosines to find $m \angle y$. The line opposite to $\angle y$ measures $80$, so now, we can set up the equation: $80^2 = 54.1^2 + 78^2 - 2(54.1)(78)$ cos $\angle y$ Evaluate exponents first, according to order of operations: $6400 = 2926.81 + 6084 - 2(54.1)(78)$ cos $\angle y$ Add to simplify on the right side of the equation: $6400 = 9010.81 - 2(54.1)(78)$ cos $\angle y$ Multiply to simplify: $6400 = 9010.81 - 8439.6$ cos $\angle y$ Subtract $9010.81$ from each side of the equation to move constants to the left side of the equation: $-2610.81 = -8439.6$ cos $\angle y$ Divide each side by $-30$: cos $\angle y = \frac{-2610.81}{-8439.6}$ Take $cos^{-1}$ to solve for $\angle y$: $m \angle y \approx 72.0^{\circ}$

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