Trignometry - SOT

62

essentially what this question says is that the height of the circumcenter from BC= inradius of the triangle

so

a/2 tan(90-A) = r = Î”/s

a/2 cot(A)=Î”/s

9

got 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) :)

1

no 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

SOT - 2