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

$m\angle A = 70^{\circ}$ $m\angle B = 110^{\circ}$ $m\angle C = 70^{\circ}$ $m\angle D = 110^{\circ}$ It is a parallelogram.
In a quadrilateral the sum of the interior angles equals to $360^{\circ}$. So: $m\angle A + m\angle B+m\angle C + m\angle D =360^{\circ}$ $x+16+2\times (x+1) + \frac{3}{2} x-11 + \frac{7}{3} x -16 = 3x +\frac{3}{2} x + \frac{7}{3} x-9=\frac{41}{6} x -9=360^{\circ}$ $\frac{41}{6} x = 369^{\circ}$ $41x= 2214^{\circ}$ $x=54^{\circ}$ Then we calculate all the angles by substituting x by $54^{\circ}$ $m\angle A = 70^{\circ}$ $m\angle B = 110^{\circ}$ $m\angle C = 70^{\circ}$ $m\angle D = 110^{\circ}$ As the adjacent angles are supplementary, this quadrilateral is a parallelogram.