1no doubt the answer will be one..
but celestine can you give a detailed solution from getting r=RcosA
and yeah , iit jee doesnot give these typ of problems nowadays
if in a triangle ABC ,the line joinning circumcentre & incentre is parallel to BC then
cosB + cosC = ???...(numerical constant).....
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3 Answers
62essentially what this question says is that the height of the circumcenter from BC= inradius of the triangle
so
a/2 tan(90-A) = r = Δ/s
a/2 cot(A)=Δ/s
9got 1
from fig
derive r =RcosA
ans follows
also try sub A=B=C=60 u easily get 1 ( this method saves a looooot of time and nowadays IITJEE dont select these type) :)
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