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# need help

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In the diagram below, A is on line segment CE and AB bisects Angle DAC. If segment DA is parallel to segment EF and angle AEF = 10angleBAC - $$22$$ degrees, then what is angle DAC in degrees? Jan 25, 2022

#1
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What is angle DAC in degrees?

Hello Guest!

$$AEF+BAC+BAD=180^o\\ 12BAC=180^o\\2BAC=\color{red}DAC=30^o\ error \\$$

$$10BAC-22^o+DAC=180^o\\ 5DAC-22^o+DAC=180^o\\ 6DAC=202^o\\ \color{blue}DAC=33\frac{2}{3}^o$$ !

Jan 25, 2022
edited by asinus  Jan 25, 2022
edited by asinus  Jan 25, 2022
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Hi Asinus! You may have misread, but angleAEF = 10angleBAC - 22 degrees, not AEF = 10angleBAC. :)

proyaop  Jan 25, 2022
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Thanks proyaop, you are right! I will correct. !

asinus  Jan 25, 2022
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By alternate interior angles, angleAEF is congruent to angleDAE.

Since CAE is a straight line, then it has a straight angle. Therefore angleBAC + angleBAD + angleDAE = 180degrees.

angleDAE = angleAEF = 10angleBAC - 22degrees.

Since angleBAC = angleBAD because BA is an angle bisector, then we can simplify the equation to:

angleBAC + angleBAC + 10angleBAC - 22 = 180degrees.

To simplify:

12 * angleBAC = 202degrees.

To simplify:

angleBAC = (101/6)degrees.

Since we are looking for angleDAC, and we also know 2 * angleBAC = angleDAC, then angleDAC = $$101\over3$$degrees. Jan 25, 2022
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∠BAC = x              ∠DAC = 2x

2x + 10x - 22 = 180

Jan 26, 2022