If #(cosA+2cosC)/(cosA+2cosB)=(sinB)/sinC# then show that the triangle is either isosceles or right angled triangle. I got #cosA+cosB+cosC=1# what to do after this?

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Answer 1 · 2018-01-18T17:19:08

#(cosA+2cosC)/(cosA+2cosB)=(sinB)/sinC#

#=>(cosA+2cosC)sinC=(cosA+2cosB)sinB#

#=>cosAsinC+2cosCsinC=cosAsinB+2cosBsinB#
#=>cosAsinC+sin2C=cosAsinB+sin2B=0#
#=>cosA(sinB-sinC)+(sin2B-sin2C)=0#

#=>cosA*2cos((B+C)/2)sin((B-C)/2)+2cos(B+C)sin(B-C)=0#

#=>cosAcos(pi/2-A/2)sin((B-C)/2)+cos(pi-A)sin(B-C)=0#

#=>cosAsin(A/2)sin((B-C)/2)-cosA*2sin((B-C)/2)cos((B-C)/2)=0#

#=>cosAsin((B-C)/2)(sin(A/2)-2cos((B-C)/2))=0#

So either #cosA=0=cos(pi/2)#

#=>A=pi/2->#triangle is right angled.

Or #sin((B-C)/2)=0#

#=>B=C-># the triangle is isoscles