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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
avatar+13890 
+1

 What is angle DAC in degrees?

 

Hello Guest!

 

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

 

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

laugh  !

 Jan 25, 2022
edited by asinus  Jan 25, 2022
edited by asinus  Jan 25, 2022
 #3
avatar+717 
+3

Hi Asinus! You may have misread, but angleAEF = 10angleBAC - 22 degrees, not AEF = 10angleBAC. :)

proyaop  Jan 25, 2022
 #4
avatar+13890 
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Thanks proyaop, you are right! I will correct.

laugh  !

asinus  Jan 25, 2022
 #2
avatar+717 
+2

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.

 

smiley

 Jan 25, 2022
 #5
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∠BAC = x              ∠DAC = 2x

2x + 10x - 22 = 180

 Jan 26, 2022

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