Triangles BDC and ACD are coplanar and isosceles. If we have angle ABC = 70, what is angle BAC, in degrees?
+336 Hello guest!
We can use the chart to help!
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 (as shown)
∠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
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-wolfie
