Question

If A, B, C are angles of a triangle and

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

triangle ABC is

• A. Right angled isosceles
• B. Isosceles
• C. Equilateral
• D. n = 0

Answer 1

Answer: (B)

Isosceles

Answer 2

R₂® R₂ – R₁

$\left| \begin{matrix} 1 & 1 & 1 \\ \sin A & \sin B & \sin C \\ \sin A + \sin^{2}A & \sin B + \sin^{2}D & \sin C + \sin^{2}C \end{matrix} \right|$R₃ ® R₃ – R₂

= $\left| \begin{matrix} 1 & 1 & 1 \\ \sin A & \sin B & \sin C \\ \sin^{2}B & \sin^{2}B & \sin^{2}C \end{matrix} \right|$

= (sin A – sin B) × (sin B – sin C) × (sin C – sin A)

Ž Either A = B or B =C or C = A So D is isosceles