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Triangles $BDC$ and $ACD$ are coplanar and isosceles. If we have $m\angle ABC = 70^\circ$, what is $m\angle BAC$, in degrees?

 Sep 16, 2017

Best Answer 

 #1
avatar+7348 
+1

Since triangle BCD is isosceles with BC = DC ,  ∠DBC = ∠BDC = 70°

 

Since ∠BDC and ∠CDA form a straight line...

∠BDC + ∠CDA  =  180°

                                               Subtract  ∠BDC  from both sides of the equation.

∠CDA  =  180° - ∠BDC

                                              ∠BDC = 70°

∠CDA  =  180° - 70°

∠CDA  =  110°

 

 

Since there are 180° in every triangle...

∠CDA + ∠DAC + ∠ACD  =  180°

                                                         ∠CDA = 110°

110° + ∠DAC + ∠ACD  =  180°

                                                         Since triangle ACD is isosceles with CD = AD,  ∠ACD = ∠DAC.

110° + ∠DAC + ∠DAC  =  180°

110° +       2∠DAC        =  180°

                 2∠DAC        =   70°

                   ∠DAC        =   35°  =  ∠BAC

 Sep 16, 2017
 #1
avatar+7348 
+1
Best Answer

Since triangle BCD is isosceles with BC = DC ,  ∠DBC = ∠BDC = 70°

 

Since ∠BDC and ∠CDA form a straight line...

∠BDC + ∠CDA  =  180°

                                               Subtract  ∠BDC  from both sides of the equation.

∠CDA  =  180° - ∠BDC

                                              ∠BDC = 70°

∠CDA  =  180° - 70°

∠CDA  =  110°

 

 

Since there are 180° in every triangle...

∠CDA + ∠DAC + ∠ACD  =  180°

                                                         ∠CDA = 110°

110° + ∠DAC + ∠ACD  =  180°

                                                         Since triangle ACD is isosceles with CD = AD,  ∠ACD = ∠DAC.

110° + ∠DAC + ∠DAC  =  180°

110° +       2∠DAC        =  180°

                 2∠DAC        =   70°

                   ∠DAC        =   35°  =  ∠BAC

hectictar Sep 16, 2017

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