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Thanks in advance!

 May 14, 2020

Best Answer 

 #1
avatar+9015 
+2

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

Since a tangent line forms a right angle with a line drawn from the center of the circle to the point of tangency:

 

m∠BCD   =   90°

 

Note that △ABC is an isosceles triangle because  BC  is the same length as  BA.

And so the base angles,  ∠BCA  and  ∠BAC ,  have the same measure.

Since the sum of the interior angles of a triangle is  180°:

 

m∠ABC  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  168°  in for  m∠ABC  since  m∠ABC  =  168°

168°  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  m∠BCA  in for  m∠BAC  since  m∠BAC  =  m∠BCA

168°  +  m∠BCA  +  m∠BCA   =   180°

                                                                    Subtract  168°  from both sides of the equation.

m∠BCA  +  m∠BCA   =   180°  -  168°

                                                                    Combine like terms.

2 * m∠BCA   =   180°  -  168°

                                                      Divide both sides of the equation by  2

m∠BCA   =   (180° - 168°) / 2

                                                      Simplify.

m∠BCA   =   6°

 

Now we can go back to the original equation and plug in the information we know:

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

m∠ACD   =   90°  -  6°

 

m∠ACD   =   84°

 May 14, 2020
 #1
avatar+9015 
+2
Best Answer

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

Since a tangent line forms a right angle with a line drawn from the center of the circle to the point of tangency:

 

m∠BCD   =   90°

 

Note that △ABC is an isosceles triangle because  BC  is the same length as  BA.

And so the base angles,  ∠BCA  and  ∠BAC ,  have the same measure.

Since the sum of the interior angles of a triangle is  180°:

 

m∠ABC  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  168°  in for  m∠ABC  since  m∠ABC  =  168°

168°  +  m∠BCA  +  m∠BAC   =   180°

                                                                    We can substitute  m∠BCA  in for  m∠BAC  since  m∠BAC  =  m∠BCA

168°  +  m∠BCA  +  m∠BCA   =   180°

                                                                    Subtract  168°  from both sides of the equation.

m∠BCA  +  m∠BCA   =   180°  -  168°

                                                                    Combine like terms.

2 * m∠BCA   =   180°  -  168°

                                                      Divide both sides of the equation by  2

m∠BCA   =   (180° - 168°) / 2

                                                      Simplify.

m∠BCA   =   6°

 

Now we can go back to the original equation and plug in the information we know:

 

m∠ACD   =   m∠BCD  -  m∠BCA

 

m∠ACD   =   90°  -  6°

 

m∠ACD   =   84°

hectictar May 14, 2020

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