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The angle bisector of an angle divides the angle into two angles with equal measure. In the diagram below, \(A\) is on segment \(\overline{CE}\) and \(\overline{AB}\) bisects \(\angle CAD\). If we have \(\overline{DA}\parallel\overline{EF}\) and \(\angle AEF\) is \(12^\circ\) less than 10 times \(\angle BAC\), then what is \(\angle CAD\) in degrees?

 Oct 26, 2017

Best Answer 

 #2
avatar+7352 
+3

 

If we look at it like this, we can see that  ∠AEF  and  ∠CAD  will add to  180° .

 

∠AEF + ∠CAD  =  180°            Plug in   10( ∠BAC ) - 12°   for  ∠AEF .

 

10( ∠BAC ) - 12°   +   ∠CAD  =  180°             And since  AB bisects ∠CAD,  ∠CAD  =  2( ∠BAC ) .

 

10( ∠BAC ) - 12°   +   2( ∠BAC )  =  180°       Add  12°  to both sides of this equation.

 

10( ∠BAC ) + 2( ∠BAC )  =  180° + 12°          Combine like terms.

 

12( ∠BAC )  =  192°                Divide both sides by  12 .

 

∠BAC  =  16°

 

And remember that

 

∠CAD  =  2( ∠BAC )      So...

 

∠CAD  =  2( 16° )

 

∠CAD  =  32°

 Oct 26, 2017
edited by hectictar  Oct 26, 2017
 #1
avatar+60 
+1

I don't know that...

 Oct 26, 2017
 #2
avatar+7352 
+3
Best Answer

 

If we look at it like this, we can see that  ∠AEF  and  ∠CAD  will add to  180° .

 

∠AEF + ∠CAD  =  180°            Plug in   10( ∠BAC ) - 12°   for  ∠AEF .

 

10( ∠BAC ) - 12°   +   ∠CAD  =  180°             And since  AB bisects ∠CAD,  ∠CAD  =  2( ∠BAC ) .

 

10( ∠BAC ) - 12°   +   2( ∠BAC )  =  180°       Add  12°  to both sides of this equation.

 

10( ∠BAC ) + 2( ∠BAC )  =  180° + 12°          Combine like terms.

 

12( ∠BAC )  =  192°                Divide both sides by  12 .

 

∠BAC  =  16°

 

And remember that

 

∠CAD  =  2( ∠BAC )      So...

 

∠CAD  =  2( 16° )

 

∠CAD  =  32°

hectictar Oct 26, 2017
edited by hectictar  Oct 26, 2017
 #3
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+2

Thanks!!!

Guest Oct 26, 2017

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