Question

If the sine of the angles of a triangle ABC satisfy the equation c³x³ - c²(a + b + c) x² + λx + μ = 0 (where a, b, c are the sides of ΔABC), then triangle ABC is

(A) always right-angled for any λ,μ

(B) right-angled only when λ = c(ab + bc + ca)/4, μ = -abc

(C) right-angled only when λ = c(ab + bc + ca)/4, μ = -abc/8

(D) never right-angled

Answer 1

Correct option (B) right-angled only when λ = c(ab + bc + ca)/4, μ = -abc

sin A + sin B + sin C = c²(a + b + c)/c³ = a + b + c/c

But sin A + sin B + sin C = a/2R + b/2R + c/2R

Comparing both we get c = 2R. So, the triangle is a right-angled triangle.

Putting the same value of c, we get λ = ab + bc + ca,μ = -abc