Tag: understanding 3d and 2d shapes

$r + R = \dfrac {a}{2}\cot \dfrac {\pi}{n} + \dfrac {a}{2}cosec \dfrac {\pi}{n}$
$= \dfrac {a}{2} \left (\dfrac {1 + \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right ) = \dfrac {a}{2} \dfrac {2\cos^{2} \dfrac {\pi}{2n}}{2\sin \dfrac {\pi}{2n}\cdot \cos \dfrac {\pi}{2n}}$
$= \dfrac {a}{2} \cot \dfrac {\pi}{2n}$.

From the figure:
Side of polygon $(AB)=1$
$AO=\dfrac { 1 }{ 2 } $
$\angle O=\dfrac { \pi  }{ 2n } $

In right angled $\triangle COA$ :
$\sin { O } =\dfrac { AC }{ AO } $
$\Rightarrow \sin { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2R } $       ..(1)

$\tan { O } =\dfrac { AC }{ CO } $
$\Rightarrow \tan { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2r } $       ...(2)

From (1) and (2)
$R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ \sin { \dfrac { \pi  }{ n }  }  } +\dfrac { 1 }{ \tan { \dfrac { \pi  }{ n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1+\cos { \dfrac { \pi  }{ n }  }  }{ \sin { \dfrac { \pi  }{ n }  }  }  \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 2\cos ^{ 2 }{ \dfrac { \pi  }{ 2n }  }  }{ 2\cos { \dfrac { \pi  }{ 2n }  } \sin { \dfrac { \pi  }{ 2n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \cot { \dfrac { \pi  }{ 2n }  } $

Ans: A

A triangle has no polygons.

Since, ${a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2}$
Using sine rule, we have
${\left(2R\sin{A}\right)}^{2}+{\left(2R\sin{B}\right)}^{2}+{\left(2R\sin{C}\right)}^{2}=8{R}^{2}$
$\Rightarrow 4{R}^{2}\left[{\left(\sin{A}\right)}^{2}+{\left(\sin{B}\right)}^{2}+{\left(\sin{C}\right)}^{2}\right]=8{R}^{2}$
$\Rightarrow {\sin}^{2}A+{\sin}^{2}B+{\sin}^{2}C=2$
$\Rightarrow 1-{\cos}^{2}A+{\sin}^{2}B+1-{\cos}^{2}C=2$
$\Rightarrow -{\cos}^{2}A+{\sin}^{2}B-{\cos}^{2}C=2-2=0$
$\Rightarrow {\cos}^{2}A-{\sin}^{2}B+{\cos}^{2}C=0$
$\Rightarrow \cos{\left(A+B\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
Since, $A+B+C=\pi \Rightarrow A+B=\pi-C$
$\Rightarrow \cos{\left(\pi-C\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  -\cos{C}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  \cos{C}\left[-cos{\left(A-B\right)}+\cos{C}\right]=0$
$\Rightarrow \cos{C}=0, -cos{\left(A-B\right)}+\cos{C}=0$
$\Rightarrow C=\dfrac{\pi}{2}$ or $-cos{\left(A-B\right)}+\cos{\left(\pi-\left(A+B\right)\right)}=0$
$\Rightarrow -cos{\left(A-B\right)}-cos{\left(A+B\right)}=0$
Using transformation angle formula, we have
$\Rightarrow 2\cos{A}\cos{B}=0$
$\therefore \cos{A}=0,\cos{B}=0$
Hence $\angle{A}=\dfrac{\pi}{2} $or $\angle{B}=\dfrac{\pi}{2}, $ or  $\angle{C}=\dfrac{\pi}{2}$

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$\left(b\right)$
$\because \sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}\le \dfrac{1}{8}$
$\Rightarrow \dfrac{r}{4R}\le \dfrac{1}{8}$
$\Rightarrow R\ge 2r$
Option$\left(c\right)$
$\because \dfrac{abc}{a+b+c}=\dfrac{4R\triangle}{2s}$
                 $=2R.r=R\left(2r\right)\le {R}^{2}$
$\therefore {R}^{2}\ge \dfrac{abc}{a+b+c}$   ($\because R\ge 2r$)
Option$\left(d\right)$
$\angle{B}={90}^{0}$
$\therefore r=\left(s-b\right)\tan{\dfrac{B}{2}}=s-b$
$R=\dfrac{b}{2\sin{B}}=\dfrac{b}{2}$
$\Rightarrow 2R=b$
$\therefore r+2R=s-b+b=s$

Option$\left(a\right)$ SInce
$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$
But here, $\tan{A}+\tan{B}+\tan{C}=0,$ is impossible.
Option$\left(b\right)$
$\dfrac{\sin{A}}{2}=\dfrac{\sin{B}}{3}=\dfrac{\sin{C}}{7}$
or $\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{5}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{c}=\dfrac{5}{7}<1$
$\therefore a+b<c$ is impossible.
Option$\left(c\right)$
${\left(a+b\right)}^{2}={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}+2ab={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=-ab$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{-1}{2}$
$\Rightarrow \cos{C}=\dfrac{-1}{2}$
$\therefore \angle{C}={120}^{0}$
and $\sqrt{2}\left(\sin{A}+\cos{A}\right)=\sqrt{3}$
$\Rightarrow \sqrt{2}\left[\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\sin{A}+\dfrac{1}{\sqrt{2}}\cos{A}\right)\right]=\sqrt{3}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow 2\left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sqrt{3}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\dfrac{\sqrt{3}}{2}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sin{\dfrac{\pi}{3}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{\pi}{3}$
$\therefore A=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$ is possible.
Option$\left(d\right)$
$\because \sin{A}+\sin{B}=\dfrac{\sqrt{3}+1}{2}$                 ................$\left(1\right)$
and $\cos{A}\cos{B}=\dfrac{\sqrt{3}}{4}=\sin{A}\sin{B}$
$\therefore \cos{A}\cos{B}-\sin{A}\sin{B}=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}=0$
$\Rightarrow \cos{\left(A+B\right)}=0$
$\Rightarrow A+B=\dfrac{\pi}{2}$
$\therefore B=\dfrac{\pi}{2}-A$
From eqn$\left(1\right)$ 
$\sin{A}+\cos{A}=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sqrt{2}\left[\sin{A}\dfrac{1}{\sqrt{2}}+\cos{A}\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{3}+1}{2}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}\left[\sin{A}\cos{\dfrac{\pi}{4}}+\cos{A}\sin{\dfrac{\pi}{4}}\right]=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\sin{\dfrac{5\pi}{12}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{5\pi}{12}$
$\Rightarrow A=\dfrac{5\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}{6}$
$\therefore B=\dfrac{\pi}{2}-A=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$
Thus, $\angle{C}=\dfrac{\pi}{2}$ is possible. 

$\because \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$
$\Rightarrow \left[\sin{C}+\sin{\left(2B+C\right)}\right]+\left[\cos{C}-\cos{\left(2B+C\right)}\right]=2\sqrt{2}$
$\Rightarrow 2\sin{\left(B+C\right)}\cos{B}+2\sin{\left(B+C\right)}\sin{B}=2\sqrt{2}$
$\Rightarrow \sin{\left(\pi-A\right)}\cos{B}+\sin{\left(\pi-A\right)}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\cos{B}+\sin{A}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\left[\cos{B}+\sin{B}\right]=\sqrt{2}$
Divide both sides by $\sqrt{2}$ we get
$\Rightarrow \sin{A}\left[\dfrac{1}{\sqrt{2}}\cos{B}+\dfrac{1}{\sqrt{2}}\sin{B}\right]=1$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$ we get
$\Rightarrow \sin{A}\left[\sin{\dfrac{\pi}{4}}\cos{B}+\cos{\dfrac{\pi}{4}}\sin{B}\right]=1$
$\Rightarrow \sin{A}\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
$\therefore \sin{A}=1$ and $\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
Hence $A={90}^{0}, \dfrac{\pi}{4}+B=\dfrac{\pi}{2}$
$\Rightarrow B=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$ (on simplification)
$\therefore A={90}^{0}, B={45}^{0}, C={45}^{0}$