Question

In $△ABC,$ which of the following statements are true:

• A. maximum value of $\sin 2A+\sin 2B+\sin 2C$ is same as the maximum value of $\sin A+\sin B+\sin C$
• B. $R\ge 2r,$ where $R$ is circumradius and $r$ is the inradius.
• C. $R^2\ge \frac{abc}{(a+b+c)}$
• D. $△ABC$ is right angled if $r+2R=s,$ where $s$ is semi perimeter.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A maximum value of $\sin 2A+\sin 2B+\sin 2C$ is same as the maximum value of $\sin A+\sin B+\sin C$
B $R\ge 2r,$ where $R$ is circumradius and $r$ is the inradius.
C $R^2\ge \frac{abc}{(a+b+c)}$
D $△ABC$ is right angled if $r+2R=s,$ where $s$ is semi perimeter.

Option$\left(a\right)$
$\because$ Maximum value of $\sin 2A+\sin 2B+\sin 2C$ and $\sin A+\sin B+\sin C$ is same that is $3$.
Option$\left(b\right)$
$\because \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\le \frac{1}{8}$
$\Rightarrow \frac{r}{4R}\le \frac{1}{8}$
$\Rightarrow R\ge 2r$
Option$\left(c\right)$
$\because \frac{abc}{a+b+c}=\frac{4R△}{2s}$
$=2R.r=R\left(2r\right)\le R^2$
$\therefore R^2\ge \frac{abc}{a+b+c}$ ($\because R\ge 2r$)
Option$\left(d\right)$
$\angle B={90}^0$
$\therefore r=(s-b)\tan \frac{B}{2}=s-b$
$R=\frac{b}{2\sin B}=\frac{b}{2}$
$\Rightarrow 2R=b$
$\therefore r+2R=s-b+b=s$