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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
avatar+315 
+2

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

If you need a more in debt answer just ask!

-wolfie

 Jan 1, 2021
 #2
avatar+1164 
+1

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
avatar+1164 
+2

the max is 44º ==> error cheeky

~~~~~~~~~~~~~~~~~~~~~~~~~~

The right answer is   ∠BAC = 80º smiley

 

https://i.imgur.com/8oM338j.png

jugoslav  Jan 2, 2021

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