Question

In AABC, AC is extended through point C to point D, mZBCD (9x – 1)', m ABC (3x + 18)°, and mZCAB (3x – 4)". Find mZCAB. - -

Answer 1

`\begin{gathered} m\angle BCD=m\angle ABC+m\angle CAB \\ 9x-1=3x+18+3x-4 \\ 9x-1=6x+14 \\ \text{solve for x:} \\ 3x=15 \\ x=(15)/(3) \\ x=5 \end{gathered}`

Since:

`\begin{gathered} m\angle CAB=3x-4 \\ m\angle CAB=3(5)-4 \\ m\angle CAB=11 \end{gathered}`

That's it. do you have any question? my pleasure