Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $R$ is the circumradius of the triangle, then

What is the value of $R^2(sin2A+sin2B+sin2C)$?

• A. $\Delta$
• B. $3\Delta$
• C. $4\Delta$
• D. $2\Delta$

Answer 1

The correct option is D $2\Delta$

$R^2(\sin {}^{2}A+\sin {}^{2}B+\sin {}^C)=R^2(\frac{a\cos A}{R}+\frac{b\cos B}{R}+\frac{c\cos C}{R})[R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}=\frac{abc}{4△}]=R(a\cos A+b\cos B+c\cos C)=\frac{abc}{4△}\times \frac{4(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^2+b^2+c^2{)}^c}{2ab}[ApplyingCosineRule]=\frac{16{△}^2}{8△}=2△$

(D) is the correct answer.