+912 
The measure of ∠BCD is 120°. The measure of ∠ABC is 85°.
What is measure of ∠BAC?
+536 The measure of ∠ACB is 60°.
The measure of ∠ABC is 85°.
...So... ∠BAC is 35°
+2446 Another option is to utilize something called the Exterior Angle Theorem. This theorem states that the measure of the exterior angle of a triangle is equal to sum of the two nonadjacent interior (also known as remote) angles. The diagram below does all the speaking for me.
Now that this theorem is established, we can save one step. In the given diagram, \(m\angle CAB+m\angle ABC=m\angle BCD\)
| \(m\angle CAB+m\angle ABC=m\angle BCD\) | Exterior Angle Theorem |
| \(m\angle CAB+85=120\) | Substitution Property of Equality |
| \(m\angle CAB=35^{\circ}\) | Subtraction Property of Equality |
Why is this the case? Well, I'm happy to show you why!
In this triangle, one is given that this is \(\triangle ABC\) with an exterior angle \(\angle BAD\). Our goal here is to prove that \(m\angle B+m\angle C=m\angle BAD\). I will utilize a two-column proof.
Let's assume that \(m\angle BAC=x^{\circ}\).
| \(m\angle BAC+m\angle B+m\angle C=180^{\circ}\) | Triangle Sum Theorem |
| \(x+m\angle B+m \angle C=180^{\circ}\) | Substitution Property of Equality |
| \(m\angle B+m\angle C=(180-x)^{\circ}\) | Subtraction Property of Equality |
| \(\angle BAC\) and \(\angle BAD\) form a linear pair | Definition of a linear pair |
| \(\angle BAC\) and \(\angle BAD\) are supplementary | \(\)Linear Pair Theorem |
| \(m\angle BAC+m\angle BAD=180^{\circ}\) | Definition of supplementary angles |
| \(x^{\circ}+m\angle BAD=180^{\circ}\) | Substitution Property of Equality |
| \(m\angle BAD=(180-x)^{\circ}\) | Subtraction Property of Equality |
| \(m\angle B+m\angle C=m\angle BAD\) | Transitive Property of Equality |
