[cavpun]

In triangle ABC If Tan A:TanB:Tanc = 1:2:3 then SinA:SinB:SinC = ?

[ebaines]

You can start with a triangle with the altitude being "d", as shown in the attached figure, and the distance from vertex A along line AB to the point of intersection with the altitude line (point P in the figure) being 1; this gives you tan(a) = d. Given that tan(b) is 2 times tan(a), this means that the distance from P to B is 1/2. You now know that sin(a) = d/sqrt(1+d^2) and sin(b) = d/sqrt(1/4 + d^2). Now you can proceed as follows:

1. Use law of sines to find sin(c) in terms of the d.
2. Use law of cosines to find cos(c), again in terms of d.
3. Divide one by the other: sin(c)/cos(c) = tan(c).
4. You were given that tan(c) is 3 times tan(a), or 3d, so set tan(c) = 3d and solve for d.
5. Shove the value for d back into the equations for sin(a), sin(b) and sin(c) as determined above, and determine the ratios to get sina:sin b:sinc.

If you do this right, you should find that d = 1.