Question

If $A,BandC$ are the angles of a triangle and
$\left|\begin{array}{ccc}1& 1& 1\\ 1+\sin A& 1+\sin B& 1+\sin C\\ \sin A+{\sin }^{2}A& \sin B+{\sin }^{2}B& \sin C+{\sin }^{2}C\end{array}\right|=0$,

then the triangle ABC is ?

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

Answer 1

The correct option is A isoceles

$\left|\begin{array}{ccc}1& 1& 1\\ 1+\sin A& 1+\sin B& 1+\sin C\\ \sin A+{\sin }^{2}A& \sin B+{\sin }^{2}B& \sin C+{\sin }^{2}C\end{array}\right|=0$

$R_2\to R_2-R_{1}and{R}_3\to R_3-R_2$

$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ \sin A& \sin B& \sin C\\ {\sin }^{2}A& {\sin }^{2}B& {\sin }^{2}C\end{array}\right|=0$

$\Rightarrow (\sin A-\sin B)(\sin B-\sin C)(\sin C-\sin A)=0$

$\therefore A=BorB=CorC=A$
Hence, triangle is an isosceles.