Regular polygons - class-VIII

$\cos{A}+\cos{B}=1$

The side length is 2. The center of the hexagon is found by rotating the vector (A2-A1) by 60 degrees. With A1=(1,0) and A2=(3,0), the center is (2, sqrt(3)). The radius is 2. The circle equation is (x-2)^2 + (y-sqrt(3))^2 = 4, which simplifies to x^2 + y^2 - 4x - 2sqrt(3)y + 17/3 = 0.

Given ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ is  a regular hexagon inscribed in a circle of unit radius
$\Rightarrow {A} _{0}{A} _{1}=1$
$\Rightarrow {A} _{0}{A} _{2}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\Rightarrow {A} _{0}{A} _{4}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore {A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}=1\times \sqrt{3}\times\sqrt{3}=3$

Area of the regular polygon will be 
$=\dfrac{nR^{2}}{2}sin(\dfrac{2\pi}{n})$.
Now 
$R=\dfrac{s}{2sin(\dfrac{\pi}{n})}$
Hence
$A=\dfrac{ns^{2}}{8sin^{2}\dfrac{\pi}{n}}.2sin(\dfrac{\pi}{n}).cos(\dfrac{\pi}{n})$

$=\dfrac{ns^{2}}{4}.cot(\dfrac{\pi}{n})$. where s is the side of the polygon.

$=nr^{2}.tan(\dfrac{\pi}{n})$ where r is the incentre.

in $\Delta{ABL}$, $AL$ is the radius of the inscribed circle. 

The area of a regular hexagon with side length 'a' is (3*sqrt(3)/2) * a^2. For a=1, Area = 1.5 * 1.732 = 2.598 cm^2.

Let $a$ be the radius of the circle 

In a regular hexagon ABCDEF, the vector CE can be found using vector addition. Since AB = a and BC = b, the vectors for the sides are related by the geometry of the hexagon, leading to CE = b - a.

The area of a regular polygon with n sides and circumradius r is given by n * (1/2 * r^2 * sin(360/n)). This formula is derived by summing the areas of n isosceles triangles with sides r and included angle 360/n.

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

Using the properties of parallel lines in a triangle, the length of the segments follows an arithmetic progression. With 35 lines, the segments divide the side into 36 equal parts; the 7th line corresponds to a ratio of 7/36 of the base, but the calculation 1.8 * (7/36) = 0.35 m is correct.

By applying Menelaus' theorem or vector geometry to the triangle and its median, a line from the vertex bisecting the median intersects the opposite side at a ratio of 1:2.

For a regular hexagon with apothem $5 m$ the side length is about $5.77m$.

Use the formula
$A = \frac{1}{2}pa$

to find the area of the hexagon.

The perimeter of the hexagon is about $6(5.77)$ or $34.62m$.

Now substitute the values.

$A = \frac{1}{2} (34.62)(5)$

Simplify.

$A = \frac{1}{2}(173.1)$

= $86.5$

$A = \frac{1}{2}(173.1)$

= $86.5$

Therefore, the area of the regular hexagon is about $86.5 m^2$.

In $\triangle ACD, \cos{C}=\dfrac{b}{\left(\dfrac{a}{2}\right)}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{2b}{a}$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2b}=2b$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=4{b}^{2}$
$\Rightarrow {a}^{2}-{c}^{2}=3{b}^{2}$

Let $A=2\alpha, B=3\alpha, C=7\alpha$
$\therefore A+B+C={180}^{0}$
$\Rightarrow 2\alpha+3\alpha+7\alpha={180}^{0}$
$\Rightarrow 12\alpha={180}^{0}$
$\Rightarrow \alpha={15}^{0}$
$\therefore A=2\alpha=2\times{15}^{0}={30}^{0}$
$B=3\alpha=3\times{15}^{0}={45}^{0}$
$C=7\alpha=7\times{15}^{0}={105}^{0}$
$a:b:c=\sin{{30}^{0}}:\sin{{45}^{0}}:\sin{{105}^{0}}$
       $=\dfrac{1}{2}:\dfrac{1}{\sqrt{2}}:\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$a:b:c=\sqrt{2}:2:\sqrt{3}+1$
        $=\dfrac{1}{\sqrt{2}}:1:\dfrac{\sqrt{3}+1}{2}$
or $1:\sqrt{2}:\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}$
or $1:\sqrt{2}:\dfrac{2\sqrt{2}}{2\left(\sqrt{3}-1\right)}$
$\therefore a:b:c= 1:\sqrt{2}:\dfrac{\sqrt{2}}{\sqrt{3}-1}$

$\cos{A}+\cos{B}+\cos{C}=\dfrac{3}{2}$
Using transformation angle formula, we have
$\left(\cos{A}+\cos{B}\right)+\cos{C}=\dfrac{3}{2}$
$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+\cos{C}=\dfrac{3}{2}$
Using sub-multiple angle formula to $\cos{C}=1-2{\sin}^{2}{\dfrac{C}{2}}$ we get
$\Rightarrow  2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}+1-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}$ 
Since $A+B+C=\pi\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{2}-\dfrac{C}{2}$
$\Rightarrow  2\cos{\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{3}{2}-1$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}-2{\sin}^{2}{\dfrac{C}{2}}=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{C}{2}\right)}\right]=\dfrac{1}{2}$
Again
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
$\Rightarrow 2\sin{\left(\dfrac{C}{2}\right)}\left[\cos{\left(\dfrac{A-B}{2}\right)}+{\cos\left(\dfrac{A+B}{2}\right)}\right]=\dfrac{1}{2}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{C}{2}\right)}\left[2\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{A}{2}\right)}\right]=\dfrac{1}{4}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}\sin{\left(\dfrac{B}{2}\right)}\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{8}$
$\therefore \sin{\left(\dfrac{A}{2}\right)}=\dfrac{1}{2}, \sin{\left(\dfrac{B}{2}\right)}=\dfrac{1}{2},\sin{\left(\dfrac{C}{2}\right)}=\dfrac{1}{2}$
$\Rightarrow \sin{\left(\dfrac{A}{2}\right)}=\sin{\dfrac{\pi}{6}}, \sin{\left(\dfrac{B}{2}\right)}=\sin{\dfrac{\pi}{6}},\sin{\left(\dfrac{C}{2}\right)}=\sin{\dfrac{\pi}{6}}$
$\therefore {\left(\dfrac{A}{2}\right)}={\left(\dfrac{B}{2}\right)}={\left(\dfrac{C}{2}\right)}=\dfrac{\pi}{6}$
$\Rightarrow \angle{A}=\angle{B}=\angle{C}=2\times\dfrac{\pi}{6}=\dfrac{\pi}{3}$
Hence, the triangle is equilateral.

${A} _{0}{A} _{1}=2.1.\cos{\dfrac{\pi}{3}}=2.1.\dfrac{1}{2}=1={A} _{1}{A} _{2}$
$\cos{\dfrac{2\pi}{3}}=\dfrac{{\left({A} _{0}{A} _{1}\right)}^{2}+{\left({A} _{1}{A} _{2}\right)}^{2}-{\left({A} _{0}{A} _{2}\right)}^{2}}{2{A} _{0}{A} _{1}.{A} _{1}{A} _{2}}$
$\Rightarrow \cos{\left(\pi-\dfrac{\pi}{3}\right)}=\dfrac{1+1-{\left({A} _{0}{A} _{2}\right)}^{2}}{2.1.1}$
$\Rightarrow -\cos{\dfrac{\pi}{3}}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow \dfrac{-1}{2}=\dfrac{2-{\left({A} _{0}{A} _{2}\right)}^{2}}{2}$
$\Rightarrow -1=2-{\left({A} _{0}{A} _{2}\right)}^{2}$
$\Rightarrow {\left({A} _{0}{A} _{2}\right)}^{2}=3$
$\therefore {A} _{0}{A} _{2}=\sqrt{3}={A} _{0}{A} _{4}$
$\therefore {A} _{0}{A} _{1}\times {A} _{0}{A} _{2}\times{A} _{0}{A} _{4}=1\times\sqrt{3}\times\sqrt{3}=3$

Inscribed circle of a regular polygon of $n$ sides
${A} _{1}=n{r}^{2}\tan{\dfrac{\pi}{n}}$
Here $n=8$
$\therefore {A} _{1}=8{r}^{2}\tan{\dfrac{\pi}{8}}$
Circumscribed circle of a regular polygon of $n$ sides is
${A} _{2}=\dfrac{n{R}^{2}}{2}\sin{\dfrac{2\pi}{n}}$
For $n=8$ we have
${A} _{2}=\dfrac{8{R}^{2}}{2}\sin{\dfrac{2\pi}{8}}$
  $=\dfrac{8{R}^{2}}{2}\sin{\dfrac{\pi}{4}}$
  $=\dfrac{8{r}^{2}}{2}2\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ (for $R=r$)
  $=8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}$ 
$\therefore \dfrac{{A} _{2}}{{A} _{1}}=\dfrac{8{r}^{2}\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{8{r}^{2}\tan{\dfrac{\pi}{8}}}$
$=\dfrac{\sin{\dfrac{\pi}{8}}\cos{\dfrac{\pi}{8}}}{\dfrac{\sin{\dfrac{\pi}{8}}}{\cos{\dfrac{\pi}{8}}}}$
$={\cos}^{2}{\dfrac{\pi}{8}}$

If radius of circle is $r$ then 
${A} _{1}{A} _{2}=2r\sin{\left(\dfrac{\pi}{n}\right)}$
${A} _{1}{A} _{3}=2r\sin{\left(\dfrac{2\pi}{n}\right)}$
${A} _{1}{A} _{4}=2r\sin{\left(\dfrac{3\pi}{n}\right)}$
$\because \dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}}$
$\Rightarrow \dfrac{1}{2r\sin{\left(\dfrac{\pi}{n}\right)}}=\dfrac{1}{2r\sin{\left(\dfrac{2\pi}{n}\right)}}+\dfrac{1}{2r\sin{\left(\dfrac{3\pi}{n}\right)}}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{3\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}+\sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\left[\sin{\left(\dfrac{3\pi}{n}\right)}-\sin{\left(\dfrac{\pi}{n}\right)}\right]=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}.2\cos{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow 2\sin{\left(\dfrac{2\pi}{n}\right)}\cos{\left(\dfrac{2\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
Using multiple angle formula, $2\sin{A}\cos{A}=\sin{2A}$ we get
$\sin{\left(\dfrac{4\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
$\therefore \dfrac{4\pi}{n}=r+{\left(-1\right)}^{r}\dfrac{3}{n}$ for $r=1,n=7$

$\because b+c=x,bc=y$
$\therefore b,c$ are the roots of ${t}^{2}-\left(b+c\right)t+bc=0$
or ${t}^{2}-xt+y=0$
$\therefore t=\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2}$
Hence, sides are  $\dfrac{x\pm\sqrt{{x}^{2}-4y}}{2},z$ 

Let $a$ be the length of side of regular polygon then
$R=\dfrac{a}{2}\csc{\left(\dfrac{\pi}{n}\right)}$ and $r=\dfrac{a}{2}\cot{\left(\dfrac{\pi}{n}\right)}$
$\therefore \dfrac{R}{r}=\dfrac{\csc{\left(\dfrac{\pi}{n}\right)}}{\cot{\left(\dfrac{\pi}{n}\right)}}=\dfrac{\dfrac{1}{\sin\left(\frac{\pi}{n}\right)}}{\dfrac{\cos{\frac{\pi}{n}}}{\sin{\frac{\pi}{n}}}}=\dfrac{1}{\cos{\left(\frac{\pi}{n}\right)}}$
$\therefore \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}$
On rationalising the denominator, we get 
$ \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1}{4}=\cos{\left(\dfrac{\pi}{5}\right)}$
On comparing $\cos{\left(\dfrac{\pi}{n}\right)}=\cos{\left(\dfrac{\pi}{5}\right)}$ we get $n=5$

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

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

The expression (b+c+a)(b+c-a) = 5bc simplifies to (b+c)^2 - a^2 = 5bc, which is b^2 + c^2 + 2bc - a^2 = 5bc, leading to b^2 + c^2 - a^2 = 3bc. By the Law of Cosines, cos(A) = (b^2 + c^2 - a^2) / (2bc) = 3bc / 2bc = 1.5. Since the cosine of an angle cannot exceed 1, no such triangle exists.

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)
$\therefore A={90}^{0}, B={45}^{0}, C={45}^{0}$