Answer:
Solution given:
CL || AM
<BCL=130°
<BAM=140°
<ABC=?
Construction:
BN || CL is made:
let
now
< BCL+<CBN=180°[shows a. property of co interior angle]
130°+p=180°
subtracting both side by 130°
130°-130°+p=180°-130°
p=50°
again
<BAM +<ABN=180°[shows property of co interior angle]
140°+q=180°
subtracting both side by 140°
140°-140°+q=180°-140°
q=40°
we have
<ABC=x=<ABN+<CBN=q+p=40°+50°=90°
Therefore the angle x=90°
