Respuesta :
We are given a triangle △ABC such that side AC is extended through C to D.
Also, we are given the following angles
[tex]\begin{gathered} m\angle BAC=6x+10 \\ m\angle ABC=6x-10 \\ m\angle BCD=8x+20 \end{gathered}[/tex]Let us first draw the triangle to better understand the problem.
As you can see, the sum of angles ∠ACD and ∠BCD must be equal to 180° as per the straight-line angle property.
[tex]\begin{gathered} m\angle ACD+m\angle BCD=180\degree \\ m\angle ACD=180\degree-m\angle BCD \\ m\angle ACD=180\degree-8x-20 \\ m\angle ACD=160\degree-8x \end{gathered}[/tex]Now recall that the sum of all three angles of a triangle must be equal to 180°
[tex]\begin{gathered} m\angle BAC+m\angle ABC+m\angle ACD=180 \\ 6x+10+6x-10+160-8x=180 \\ 6x+6x+160-8x=180 \\ 12x-8x=180-160 \\ 4x=20 \\ x=\frac{20}{4} \\ x=5 \end{gathered}[/tex]Therefore, the value of x is 5
