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IN A TRIANGLE ABC , sides a:b:c = 4:5:6 ,
then ratio of circumradius to inradius is???
Given
a:b:c = 4:5:6
so a=4k, b=5k, c=6k
we know cosA=b^2 + c^2 - a^2
2bc
on solving we get
cosA=3/4 so (sinA)^2=7/16------(1) <(sinA)^2=1-(cosA)^2>
circumradius, R=abc/4*area of triangle -----------(2)
inradius, r=area of triangle/semi perimeter of triangle -----(3)
area of triangle= bc*sinA/2 = (15k^2)sinA ---------(4)
semi perimeter, s=a+b+c/2 = 15k/2 -----------(5)
we are required to find R/r
on solving (2) and (3) we get
R/r = s*a/bc(sinA)^2
using (1), (4), (5)
we get
R/r = 16/7
hi Rahul
PFA this will definetly help
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