The correct option is D sinα+sinβ+sinγcosα+cosβ+cosγ
Given vertices of the triangle are A(x1,x1tanα),B(x2,x2tanβ) and C(x3,x2tanγ)
Assuming S,G, and H be circumcente, centroid and orthocentre of △ABC respectively, then
S≡(0,0)G≡(x1+x2+x33,x1tanα+x2tanβ+x3tanγ3)H≡(a,b)
We know that, centroid divide circumcentre and orthocentre in the ratio 1:2
x1+x2+x33=(0×2)+(a×1)3⇒a=x1+x2+x3x1tanα+x2tanβ+x3tanγ3=(2×0)+(1×b)3⇒b=x1tanα+x2tanβ+x3tanγ
As S is the circumcentre, so
AS=BS=CS=r√(x1−0)2+(x1tanα−0)2=r⇒x1secα=r⇒x1=rcosα
Similarly,
x2=rcosβx3=rcosγ
Now,
ba=x1tanα+x2tanβ+x3tanγx1+x2+x3
Putting x1=rcosα,x2=rcosβ,x3=rcosγ, we get
∴ba=sinα+sinβ+sinγcosα+cosβ+cosγ