Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 4 - Section 4.2 - The Parallelogram and Kite - Exercises - Page 186: 33


by Theorem one diagonal of a kite bisects two angles of the kite which is $\angle A and \angle C $ therefore $\triangle ABC = \frac{1}{2} \angle A + \frac{1}{2} \angle C + \angle B= 180^{\circ}$ By substitution $ (\angle B - 30) + (\angle B -50) + \angle B = 180 $ $So \angle B = 110$

Work Step by Step

In the kite $ABCD$, the measures of $\angle B$ and $\angle D$ are equal. Also, the sum of the four interior angles in a kite is $360^{\circ}$ We can find $m\angle B$: $m\angle A + m\angle B + m\angle C + m\angle D = 360^{\circ}$ $(m\angle B - 50) + m\angle B + (m\angle B-30) + m\angle B = 360^{\circ}$ $4~m\angle B - 80 = 360^{\circ}$ $4~m\angle B = 440^{\circ}$ $m\angle B = \frac{440^{\circ}}{4}$ $m\angle B = 110^{\circ}$
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