Arun
Last Activity: 5 Years ago
Here angle ACB=120 and angle CAB=40 degreeSo Angle ABC= 180−120−40−20 degreeand using linear pair property, ∠BCP=180−120=60 degreeNow lets take mid point of CP as M and then CM=MP........(1)GivenAP=AC+2CBor AC+CP=AC+2CBCP=2CB........(2)CM+MP=2CBCM+CM=2CB2CM=2CBCM=CBmeans triangle CMB is isosceles, So in triangle BMC, ∠CMB=∠CBMUsing angle sum property in triangle BMC∠CMB+∠CBM=180−60=120or ∠CMB=∠CBM=60 degreeSo angle BMP= 180−60=120 degree [Linear pair]So finally triangle BMCis equilateral triangle.So CM=MBNow using equation 1, MP=MBSo in triangle BMP, ∠MBP=∠MPB.............(3)Using angle sum property in triangle MPB, ∠MBP+∠MPB+120=180or ∠MBP+∠MBP=60∠MBP=30 degreeSo angle ABP=20+60+30=110 degree
