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In a △ABC,a=a1=2,b=a2,c=a3 such that ap+1=5p32−pap(22−p−4p−25pap), where p=1,2 and r1,r2,r3 are ex-radii, then r3r1 is

Answer» In a ABC,a=a1=2,b=a2,c=a3 such that ap+1=5p32pap(22p4p25pap), where p=1,2 and r1,r2,r3 are ex-radii, then r3r1 is


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