R1/BC+r2/CA+r3/ab=1/r-1/2r
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Answer:
Hence proved that (r1/bc)+(r2/ca)+(r3/ab)=1/r-1/2R
Step-by-step explanation:
r1/bc + r2/ca + r3/ab = 1/abc[ar1 + br2 + cr3]
= 1/abcΣar1 = 1/abcΣa s tanA/2
= s/abc Σ2RsinAtanA/2
= s4R/abcΣsinA/2cosA/2Sin/Cos(A/2)/(A/2)
= s 1/Δ Σsin^2A/2
= 1/r Σ 1 - cos A/2
= 1/2r [ 1 - cosA + 1 - cosB + 1 - cosC ]
= 1/2r [ 3 - ( cos A + cos B + cos C)]
= 1/2r [ 3 - ( 1+ 4sinA/2sinB/2sinC/2)]
= 1/2r [ 2 -4RsinA/2sinB/2sinc/2/R]
= 1/2r [ 2 - r/R]
= 1/r - 1/2R
Hence proved that (r1/bc)+(r2/ca)+(r3/ab)=1/r-1/2R
Step-by-step explanation:
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