[tex]\angle{DBC }[/tex]=(5x+4)°
[tex]\angle{ A }[/tex]=(2x-2)°
[tex]\angle{C }[/tex]=(4x-6)°
we know that,
[tex]\angle{DBC}[/tex]=[tex]\angle{A}[/tex]+[tex]\angle{C}[/tex]
According to the question,
[tex]\tt{ 5x+4=2x-2+4x-6 }[/tex]
[tex]\tt{ 5x+4=6x-8 }[/tex]
[tex]\tt{ 6x-5x=4+8 }[/tex]
[tex]\blue{x=12 }[/tex]
so,
[tex]\angle{C }[/tex]=(4x-6)°=(4×12-6)°=(48-6)°=42°
