Question

If the system of equations $x\sin \alpha +y\sin \beta +x\sin \gamma =0,$ $x\cos \alpha +y\cos \beta +z\cos \gamma =0$, $x+y+z=0$ where $\alpha ,\beta ,\gamma$ are angles of a triangle, have a non-trivial solution, then the triangle must be

• A. isosceles
• B. equilateral
• C. right- angled
• D. none of these

Answer 1

The correct option is A isosceles

Since the given system has a non-trivial solution, therefore,

$\left|\begin{array}{ccc}\sin \alpha & \sin \beta & \sin \gamma \\ \cos \alpha & \cos \beta & \cos \gamma \\ 1& 1& 1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}\sin \alpha & \sin \beta -\sin \alpha & \sin \gamma -\sin \alpha \\ \cos \alpha & \cos \beta -\cos \alpha & \cos \gamma -\cos \alpha \\ 1& 0& 0\end{array}\right|=0$

$[$Applying $C_2\to C_2-C_1,C_3\to C_3-C_1]$

$\Rightarrow \left|\begin{array}{ccc}\sin \alpha & 2\cos \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\beta -\alpha }{2}\right)& 2\cos \left(\frac{\gamma +\alpha }{2}\right)\sin \left(\frac{\gamma -\alpha }{2}\right)\\ \cos \alpha & 2\sin \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)& 2\sin \left(\frac{\gamma +\alpha }{2}\right)\sin \left(\frac{\alpha -\gamma }{2}\right)\\ 1& 0& 0\end{array}\right|=0$

$\Rightarrow 4\sin \left(\frac{\alpha -\beta }{2}\right)\sin \left(\frac{\gamma -\alpha }{2}\right)$

$[\sin \left(\frac{\gamma +\alpha }{2}\right)\cos \left(\frac{\beta +\alpha }{2}\right)-\sin \left(\frac{\beta +\alpha }{2}\right)\cos \left(\frac{\gamma +\alpha }{2}\right)]=0$

$\Rightarrow -4\sin \left(\frac{\alpha -\beta }{2}\right)\sin \left(\frac{\beta -\gamma }{2}\right)\sin \left(\frac{\gamma -\alpha }{2}\right)=0$

$\Rightarrow \sin \left(\frac{\alpha -\beta }{2}\right)=0$

or $\sin \left(\frac{\beta -\gamma }{2}\right)=0$ or $\sin \left(\frac{\gamma -\alpha }{2}\right)=0$

$\Rightarrow \alpha -\beta =0$ or $\beta -\gamma =0$ or $\gamma -\alpha =0$

$\Rightarrow \alpha =\beta$ or $\beta =\gamma$ or $\gamma =\alpha$

$\Rightarrow$ triangle must be isosceles.