Question

If the line $\begin{array}{c}x{\sin }^{2}A+y\sin A+1=0\\ x{\sin }^{2}B+y\mathrm{sinB}+1=0\\ x{\sin }^{2}C+y\mathrm{sinC}+1=0\end{array}$ are concurrent where $A,B,C$ are angles of triangle then $\Delta ABC$ must be

• A. equilateral
• B. isosceles
• C. right angle
• D. no such triangle exist

Answer 1

Answer: (A)

equilateral

$x{\sin }^{2}A+y\sin A+1=0$

$x{\sin }^{2}B+y\sin B+1=0$

$x{\sin }^{2}C+y\sin C+1=0$

are concurrent then

$\frac{{\sin }^{2}A}{\sin A}=\frac{{\sin }^{2}B}{\sin B}=\frac{{\sin }^{2}C}{\sin C}$

$\sin A=\sin B=\sin C$

$A=B=C$

Triangle is equilateral

$A$ is correct.