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In ΔABC ∠ACB = 120º and ∠CAB = 40º. AC is extended to P such that AP = AC + 2CB. The measure of ∠ABP is 1) 60º 2) 120º 3) 110º 4) 100º

In ΔABC ∠ACB = 120º and ∠CAB = 40º. AC is extended to P such that AP = AC + 2CB. The measure of ∠ABP is
 
1) 60º
2) 120º
3) 110º
4) 100º 

Grade:7

2 Answers

Arun
25750 Points
4 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
 
Rahul Sanapala
44 Points
4 years ago
Seems like there is a problem. Your answer/solution is blank, could you please kindly repost the solution , Sir?

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