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# triangle question

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

Jan 1, 2021

#1
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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

-wolfie

Jan 1, 2021
#2
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Triangles BDC and ACD are coplanar and isosceles. If we have angle ABC = 70, what is angle BAC, in degrees?

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The diagram shows that the angle ABC cannot be 70º; the max is 44º (as integer)

Jan 1, 2021
#3
+1164
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the max is 44º ==> error

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The right answer is   ∠BAC = 80º

jugoslav  Jan 2, 2021