Answer:
\angle ACB=40^{\circ}, \angle ABC=63^{\circ}
Step-by-step explanation:
Finding x:
We know that [tex]\angle DAB=180^{\circ}[/tex], because it's a straight line. So, [tex]\angle CAB=180^{\circ}-103^{\circ}=77^{\circ}[/tex].
We know that the sum of the angles of any triangle will be [tex]180^{\circ}[/tex].
So, we have that [tex](3x-14)+(3x+9)+77=180[/tex].
Combining like terms on the left gives [tex]6x+72=180[/tex].
Subtracting [tex]72[/tex] from both sides gives [tex]6x=108[/tex].
Dividing both sides by [tex]6[/tex] gives [tex]x=18[/tex].
Finding missing angle measures:
[tex]\angle ACB=3x-14=3(18)-14=54-14=40[/tex].
[tex]\angle ABC=3x+9=3(18)+9=54+9=63[/tex].
So, [tex]\angle ACB=40^{\circ}[/tex] and [tex]\angle ABC=63^{\circ}[/tex] and we're done!
Note: [tex]\angle ACB+\angle ABC=\angle CAD[/tex]. Coincidence? If not, try and prove it!
