Question

If $\alpha ,\beta ,\gamma$ are the angles of a triangle and the 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

Answer 1

Answer: (B)

isosceles

Given system of equations has non-trivial solution.
$\Rightarrow \left|\begin{array}{ccc}\cos (\alpha -\beta )& \cos (\beta -\gamma )& \cos (\gamma -\alpha )\\ \cos (\alpha +\beta )& \cos (\beta +\gamma )& \cos (\gamma +\alpha )\\ \sin (\alpha +\beta )& \sin (\beta +\gamma )& \sin (\gamma +\alpha )\end{array}\right|=0$

$\Rightarrow \cos (\alpha -\beta )\sin (\alpha -\beta )+\cos (\beta -\gamma )\sin (\beta -\gamma )+\cos (\gamma -\alpha )\sin (\gamma -\alpha )=0$

$\Rightarrow \sin (2\alpha -2\beta )+\sin (2\beta -2\gamma )+\sin (2\gamma -2\alpha )=0$

As $(2\alpha -2\beta )+(2\beta -2\gamma )+(2\gamma -2\alpha )=0$
$4\sin (\alpha -\beta )\sin (\beta -\gamma )\sin (\gamma -\alpha )=0$
$\Rightarrow \alpha =\beta$ or $\beta =\gamma$ or $\gamma =\alpha$
$\therefore$ Triangle is necessarily an Isosceles.
Hence, option B.