The ratio of angles in a triangle `ABC` is `2:3:7` then prove that `a:b:c=sqrt2:2:(sqrt3+1)`

The ratio of angles in a triangle ABC is 2:3:7 then prove that a:b:c=sqrt(2):2:(sqrt(3)+1)

The ratio of the sides of a triangle ABC is 1:sqrt(3):2. The ratio A:B:C is

The angles of a triangle ABC are in A.P .and it is being given that b:c=sqrt(3):sqrt(2), find /_A.

The angles of a triangle ABC are in AP and if it is being given the b:c=sqrt3:sqrt2,"find"/_A, /_Band/_C.

In a triangle ABC if 2a=sqrt(3)b+c , then possible relation is

If in a triangle ABC, (tanA)/1= (tanB)/2 = (tanC)/3 then prove that 6sqrt(2a)=3sqrt(5b)=2sqrt(10)c

If A(1,-1,2),B(2,1,-1)C(3,-1,2) are the vertices of a triangle then the area of triangle ABC is (A) sqrt(12) (B) sqrt(3) (C) sqrt(5) (D) sqrt(13)

The equation of the base of an equilateral triangle ABC is x+y=2 and the vertex is (2,-1). The area of the triangle ABC is: (sqrt(2))/(6) (b) (sqrt(3))/(6) (c) (sqrt(3))/(8) (d) None of these

In a triangle ABC,if (sqrt(3)-1)a=2b,A=3B, then /_C is