Answer:[tex]70^{\circ}[/tex]
Step-by-step explanation:
Given
[tex]\angle BAC=112^{\circ}\\\angle BCM=(2x+12)^{\circ}\\\angle ABL=2x^{\circ}[/tex]
[tex]\angle ACB[/tex] is given by [tex]180^{\circ}-\angle BCM[/tex] as the angle around a straight line is [tex]180^{\circ}[/tex]
[tex]\therefore \angle BCA=180^{\circ}-2x^{\circ}-12^{\circ}\\\angle BCA=(168-2x)^{\circ}[/tex]
Also, using exterior angle property. We can write
[tex]\angle BCA+\angle BAC=\angle ABL[/tex]
Insert the values
[tex]\Rightarrow 168^{\circ}-2x^{\circ}+112^{\circ}=2x^{\circ}\\\Rightarrow 280=4x^{\circ}\\\Rightarrow x=70^{\circ}[/tex]
