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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?

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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

#1
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+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
+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

hectictar Sep 16, 2017