## Geometry: Common Core (15th Edition)

$m \angle x \approx 36.9^{\circ}$ $m \angle y \approx 53.1^{\circ}$
First, we want to know what side is opposite the angle in question. The side that is opposite to the angle we are looking for, $\angle x$, has a measure of $3$, so let's plug in what we know into the formula for the law of cosines: $3^2 = 4^2 + 5^2 - 2(4)(5)$ cos $\angle x$ Evaluate exponents first, according to order of operations: $9 = 16 + 25 - 2(4)(5)$ cos $\angle x$ Add to simplify on the right side of the equation: $9 = 41 - 2(4)(5)$ cos $\angle x$ Multiply to simplify: $9 = 41 - 40$ cos $\angle x$ Subtract $41$ from each side of the equation to move constants to the left side of the equation: $-32= -40$ cos $m \angle x$ Divide each side by $-40$: cos $m \angle x = \frac{-32}{-40}$ Take $cos^{-1}$ to solve for $\angle x$: $m \angle x \approx 36.9^{\circ}$ Now, we use the law of cosines to find $m \angle y$. The line opposite to $\angle y$ measures $4$, so now, we can set up the equation: $4^2 = 3^2 + 5^2 - 2(3)(5)$ cos $\angle y$ Evaluate exponents first, according to order of operations: $16 = 9 + 25 - 2(3)(5)$ cos $\angle y$ Add to simplify on the right side of the equation: $16 = 34 - 2(3)(5)$ cos $\angle y$ Multiply to simplify: $16 = 34 - 30$ cos $\angle y$ Subtract $34$ from each side of the equation to move constants to the left side of the equation: $-18 = -30$ cos $m \angle y$ Divide each side by $-30$: cos $m \angle y = \frac{-18}{-30}$ Take $cos^{-1}$ to solve for $\angle y$: $m \angle y \approx 53.1^{\circ}$