  Question

# The vertices of an acute angled triangle are A(x1,x1tanα),B(x2,x2tanβ) and C(x3,x2tanγ). If origin is the circumcentre of △ABC and H(a,b) be its orthocentre, then ba equals to (where x1,x2,x3 are positive)cosα+cosβ+cosγcosαcosβcosγsinα+sinβ+sinγsinαsinβsinγsinα+sinβ+sinγcosα+cosβ+cosγtanα+tanβ+tanγtanαtanβtanγ

Solution

## The correct option is C sinα+sinβ+sinγcosα+cosβ+cosγGiven 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  Suggest corrections   