Please slove question 37
Q.37. If one angle of a triangle is 110 ° , then the angle between the bisectors of exterior angles of the two angles is:
(A) 55 °
(B) 65 °
(C) 45 °
(D) 35 °

Dear Student ,
Please find below the solution to the asked query :

Here BP and CP are the angle bisectors of exterior angles CBD and BCESo,CBP=CBD2 and BCP=BCE2A=110°       GivenIn ABC,A+B+C=180°  ...1   Angle Sum PropertyCBD=A+C    ... 2BCE=A+B   ... 3Adding 2 and 3 , we getCBD + BCE= A+A +B+CCBD + BCE= 110°+180°                         Using 1 CBD + BCE= 290°Now,In BCPCBP+BCP+BPC=180°         Angle Sum PropertyCBD2+BCE2+BPC=180°CBD + BCE2+BPC=180°290°2+BPC=180°145°+BPC=180°BPC=35°     ANS...
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