1.

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 necessarilya. equilateral b. isoscelesc. right angled`""`d. acute angledA. equiliateralB. isosocelesC. right angledD. acute angled

Answer» Correct Answer - B
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.


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