Question

If A,B,C are the angles of triangle ABC, then $\left|\begin{array}{ccc}sin2A& sinC& sinB\\ sinC& sin2B& sinA\\ sinB& sinA& sin2C\end{array}\right|$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{3\sqrt{3}}{8}$

Answer 1

Answer: (B)

$0$

$ABC$ is a triangle

$\left[\begin{array}{ccc}\sin 2A& \sin C& \sin B\\ \sin C& \sin 2B& \sin A\\ \sin B& \sin A& \sin 2C\end{array}\right]$

$ABC$ is a triangle

$\Rightarrow$ It may also be equilateral

$\Rightarrow A=B=C=60$

$\Rightarrow \left[\begin{array}{ccc}\sin 120& \sin 60& \sin 60\\ \sin 60& \sin 120& \sin 60\\ \sin 60& \sin 60& \sin 120\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}& \frac{\sqrt{3}}{2}\end{array}\right]$

$=0$

$\therefore det=0$