In #DeltaABC#, the value of #(a cosA + b cosB + c cosC)/(a+b+c)# is equal to?

Question

A)#r/R#
B)#R/(2r)#
C)#R/r#
D)#(2r)/R#

where r = inradius and R = circumradius

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Answer 1 · 2017-12-09T03:33:03

Using sine rule #a/sinA=b/sinB=c/sinC=2R#
Numerator of the given expression
#=acosA+bcosB+c cosC#

#=acosA+2RsinBcosB+2RsinCcosC#

#=acosA+R(2sinBcosB+2sinCcosC)#

#=acosA+R(sin2B+sin2C)#

#=acosA+2Rsin(B+C)cos(B-C)#

#=acos(pi-(B+C))+2Rsin(pi-A)cos(B-C)#

#=-acos(B+C)+2RsinAcos(B-C)#

#=-acos(B+C)+acos(B-C)#

#=a(cos(B-C)-cos(B+C))#

#=2asinBsinC#

Now inserting #sinB=(2Delta)/(ac) andsinC=(2Delta)/(ab) #

#=2axx(2Delta)/(ac)xx(2Delta)/(ab)#

#=(8Delta^2)/(abc)#

#=(8Delta^2)/(4RDelta)# #" "color (red)("[since "(abc)/(4R)=Delta]#

#=(2Delta)/R#

Again we know

#1/2r(a+b+c)=Delta#

So denominator of the given expression becomes

#a+b+c=(2Delta)/r#

Hence the given expression

#=(acosA+bcosB+c cosC)/(a+b+c)#

#=((2Delta)/R)/((2Delta)/r)#

#=r/R#