If `alpha,beta,gamma` are the angles of a triangle and system of equations `cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0` `cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0` `sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0` has non-trivial solutions, then triangle is necessarily a. equilateral b. isosceles c. right angled`""` d. acute angled

Options

A. equiliateral
B. isosoceles
C. right angled
D. acute angled

Answer

B

Explanation

Let `Delta = |{:(cos (alpha -beta),,cos (beta-gamma),,cos (gamma-alpha)),(coa(alpha+beta),,cos(beta+gamma),,cos(gamma+alpha)),(sin(alpha+beta),,sin(beta+gamma),,sin(gamma+alpha)):}|`
It is clear that either `alpha =beta " or " beta = gamma " or " gamma =alpha ` is suffieicient to make `Delta =0 ` . It is not necessary that triangle is equilateral.
Also isosceles traingle can be obtuse one.