IN any triangle ABC,#sinA-cosB=cosC# then angle B is?

options given are

  1. #pi/2#
  2. #pi/3#
  3. #pi/4#
  4. #pi/6#

1 Answer
maganbhai P.
Aug 10, 2018

#1 .B=pi/2#

Explanation:

We know that,

#(1)cosx+cosy=2cos((x+y)/2)cos((x-y)/2)#

#(2)A+B+C=pi=A=pi-(B+C)# #to[becausetriangleABC]#

#(3)cosx-cosy=-2sin((x+y)/2)sin((x-y)/2)#

Using #(1) and(2)#

#sinA=cosB+cosC#

#:.sin[pi-(B+C)]=2cos((B+C)/2)cos((B-C)/2)#

#:.sin(B+C)=2cos((B+C)/2)cos((B-C)/2)#

Using #sin2theta=2sinthetacostheta, #we get

#2sin((B+C)/2)cos((B+C)/2)=2cos((B+C)/2)cos((B-C)/2)#

#:.sin((B+C)/2)=cos((B-C)/2)to[becausecos((B+C)/2)!=0]#

#cos[pi/2-(B+C)/2]#=#cos((B-C)/2)to[becausecos(pi/2- x)#=#sinx]#

#cos[pi/2-(B+C)/2]-cos((B-C)/2)=0#

Using #(3)#

#-2sin((pi/2-(B+C)/2+(B-C)/2)/2)sin((pi/2-(B+C)/2-(B-C)/2)/2)=0#

Simplifying we get

#sin(pi/4-C/2)sin(pi/4-B/2)=0#

#sin(pi/4-C/2)=0 or sin(pi/4-B/2)=0#

#=>pi/4-C/2=0 or pi/4-B/2=0#

#=>C/2=pi/4 or B/2=pi/4#

#=>C=pi/2 orB=pi/2#

Hence , #B=pi/2#

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