## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 3 - Review Exercises - Page 166: 22

#### Answer

m$\angle$DAC=115$^{\circ}$

#### Work Step by Step

Step one : since the triangle ABC is an isosceles triangle so that the base angles are congruent which means $180=m\angle B + \angle A+m \angle B$ By substitution $180=50+2\angle A$ So $\angle A=\angle B= 65$. Step two : $\angle C= \angle BCD + \angle DCA = 65^{\circ} and \angle A= \angle BAD + \angle DAC = 65^{\circ}$ since angle A equal to Angle C and given Angle BCD equal to angle DAC therefore, we conclude that $\angle DCA= \angle BAD$ . Step three : by constructing an exterior angle of a triangle ADC and we know that the exterior angle of a triangle equal the sum measures of the two non adjacent interior angles. The exterior angle of triangle ADC is $115 + \angle BCD =\angle ADC + \angle DAC$ Since its given that angle BCD is congruent to angle DAC , so by substitution $\angle ADC = 115^{\circ}$.

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