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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)
  1. cosα+cosβ+cosγcosαcosβcosγ
  2. sinα+sinβ+sinγsinαsinβsinγ
  3. sinα+sinβ+sinγcosα+cosβ+cosγ
  4. tanα+tanβ+tanγtanαtanβtanγ


Solution

The correct option is C 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)3a=x1+x2+x3x1tanα+x2tanβ+x3tanγ3=(2×0)+(1×b)3b=x1tanα+x2tanβ+x3tanγ

As S is the circumcentre, so
AS=BS=CS=r(x10)2+(x1tanα0)2=rx1secα=rx1=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

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