Geometry: Common Core (15th Edition)

$m \angle 1 = 62^{\circ}$ $m \angle 2 = 118^{\circ}$ $m \angle 3 = 118^{\circ}$ $x = \frac{5}{2}$
In an isosceles trapezoid, base angles are congruent to one another; therefore, $m \angle 1 = 62^{\circ}$. In a trapezoid, interior angles add up to $360^{\circ}$. Let's set up an equation to find $m \angle 2$ and $m \angle 3$: $62^{\circ} + 62^{\circ} + m \angle 2 + m \angle 3 = 360^{\circ}$ Add constants on the left side of the equation to simplify: $124^{\circ} + m \angle 2 + m \angle 3 = 360^{\circ}$ Subtract $124^{\circ}$ from each side of the equation to move constants to the right side of the equation: $m \angle 2 + m \angle 3 = 236^{\circ}$ Again, base angles of an isosceles trapezoid are congruent; therefore, $m \angle 2$ is half of $236^{\circ}$, and $m \angle 3$ is also half of $236^{\circ}$: $m \angle 2 = m \angle 3 = 236^{\circ} \div 2$ Divide to solve: $m \angle 2 = m \angle 3 = 118^{\circ}$ The legs of an isosceles trapezoid are congruent to one another, so let us set the two legs equal to one another: $x + 2 = 5x - 8$ Subtract $2$ from each side of the equation to move constants to the right side of the equation: $x = 5x - 10$ Subtract $5x$ from each side of the equation to move variable terms to the left side of the equation: $-4x = -10$ Divide both sides by $-4$ to solve for $x$: $x = \frac{10}{4}$ Divide both the numerator and denominator by their greatest common factor: $x = \frac{5}{2}$