Properties of Triangles

$ \begin{aligned} & \text { Given, }\left|\begin{array}{lll} b & 1 & a \\ a & 1 & c \\ c & 1 & b \end{array}\right|=0 \\ & \Rightarrow b(b-c)-1\left(a b-c^2\right)+a(a-c)=0 \\ & \Rightarrow \quad b^2-b c-a b+c^2+a^2-a c=0 \\ & \Rightarrow \quad a^2+b^2+c^2-a b-b c-a c=0 \\ & \Rightarrow \quad(a-b)^2+(b-c)^2+(c-a)^2=0 \\ & \Rightarrow \quad a=b=c \text { i.e } A=B=C=60^{\circ} \end{aligned} $

[ ∵ Equilateral triangle has 3 sides of equal length and each angle is of measure $60^{\circ}$ ]

$ \begin{aligned} \therefore \quad 2(\cos A+ & \cos B+\cos C) \\ & =2 \times 3 \times \cos 60^{\circ}=3 \end{aligned} $

In $\triangle A B C, A=\frac{\pi}{3}, B=\frac{\pi}{4}$

We know by angle sum property,

$ \begin{aligned} & & A+B+C & =\pi \\ \Rightarrow & & \frac{\pi}{3}+\frac{\pi}{4}+C & =\pi \\ \Rightarrow & & C & =\pi-\frac{7 \pi}{12}=\frac{5 \pi}{12} \end{aligned} $

Using sine rule,

$ \begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ \Rightarrow & \frac{a}{\frac{\sqrt{3}}{2}}=\frac{b}{\frac{1}{\sqrt{2}}}=\frac{c}{\frac{\sqrt{3}+1}{2 \sqrt{2}}}=k \\ \Rightarrow & a=\frac{\frac{\sqrt{3}}{2}}{2} k, b=\frac{k}{\sqrt{2}}, c=\frac{(\sqrt{3}+1)}{2 \sqrt{2}} k \\ \therefore & \frac{a^2-b^2}{c^2}=\frac{\frac{3}{4} k^2-\frac{1}{2} k^2}{\frac{4+2 \sqrt{3}}{8} k^2} \end{aligned} $

                    $ \begin{aligned} & =\frac{\frac{1}{4}}{\frac{2+\sqrt{3}}{4}} \\ & =\frac{1}{2+\sqrt{3}}=2-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \text { In } \triangle A B C, a=3, b=7, c=8 \text {, } \\ & s=\frac{a+b+c}{2}=9 \end{aligned}\\ &\text { Now, }\\ &\begin{aligned} \sin \frac{B}{2} \tan & \frac{C-A}{2}=\sin \frac{B}{2} \cdot \frac{c-a}{c+a} \cdot \cot \frac{B}{2} \\ & =\frac{c-a}{c+a} \cos \frac{B}{2} \\ & =\frac{c-a}{c+a} \sqrt{\frac{s(s-b)}{a c}} \\ & =\frac{8-3}{8+3} \sqrt{\frac{9 \times(9-7)}{3 \times 8}}=\frac{5 \sqrt{3}}{22} \end{aligned} \end{aligned} $

Let $B C, A C$ and $A B$ are $a, b$ and $c$ respectively.

In $\triangle A B D$,

$G$ is the centroid and centroid divides the median in the ratio $2: 1$

$ \begin{array}{ll} A G=\frac{2}{3} \times 4=\frac{8}{3} \\ \text { In } \triangle A B G, \frac{\sin \frac{\pi}{6}}{B G}=\frac{\sin \frac{\pi}{3}}{A G} \\ \Rightarrow \frac{1}{2 B G}=\frac{\sqrt{3}}{2 \times \frac{8}{3}} \\ \Rightarrow B G=\frac{8}{3 \sqrt{3}}=\frac{8 \sqrt{3}}{9} \end{array} $

In $\triangle A G B$,

$ \begin{aligned} & \frac{\pi}{6}+\frac{\pi}{3}+\angle A G B=\pi \Rightarrow \angle A G B=\frac{\pi}{2} \\ \therefore & \text { Area of } \triangle A G B=\frac{1}{2} \times A G \times B G \\ = & \frac{1}{2} \times \frac{8}{3} \times \frac{8 \sqrt{3}}{9}=\frac{32 \sqrt{3}}{27} \end{aligned} $

As the median divides the triangle in equal parts

Area of $\triangle A B C=3$ (Area of $\triangle A G B$ )

$ \begin{aligned} & =3 \times \frac{32 \sqrt{3}}{27}=\frac{32 \sqrt{3}}{9} \text { sq. units } \\ & =\frac{32}{3 \sqrt{3}} \text { sq. units } \end{aligned} $

$ \begin{aligned} &\begin{aligned} \cos C & =\frac{a^2+b^2-c^2}{2 a b}=\frac{25+36-81}{2 \times 5 \times 6} \\ & =-\frac{20}{60}=-\frac{1}{3} \end{aligned}\\ &C \text { is an obtuse angle }\\ &\begin{aligned} \sin C & =\sqrt{1-\cos ^2 C}=\sqrt{1-\left(\frac{1}{9}\right)}=\frac{2 \sqrt{2}}{3} \\ \sin 2 C & =2 \sin C \cos C \\ & =2 \times \frac{2 \sqrt{2}}{3} \times\left(-\frac{1}{3}\right)=\frac{-4 \sqrt{2}}{9} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } \frac{r}{r_1}=\frac{1}{2} \\ & \begin{array}{l} \frac{4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}=\frac{1}{2} \\ \tan \frac{B}{2} \tan \frac{C}{2}=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ A+B+C=\pi \\ \Rightarrow \quad \tan \left(\frac{B}{2}+\frac{C}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{A}{2}\right) \\ \Rightarrow \quad \frac{\tan \frac{B}{2}+\tan \frac{C}{2}}{1-\tan \frac{B}{2} \tan \frac{C}{2}}=\cot \frac{A}{2} \\ \Rightarrow \quad \frac{\tan \frac{B}{2}+\tan \frac{C}{2}}{1-\frac{1}{2}}=\frac{1}{\tan \frac{A}{2}} \\ \quad[\because \text { From Eq. (i) }] \\ \quad 2 \tan \frac{A}{2}\left[\tan \frac{B}{2}+\tan \frac{C}{2}\right]=1 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Multiply by } 2 \text { on both sides, }\\ &4 \tan \frac{A}{2}\left[\tan \frac{B}{2} \tan \frac{C}{2}\right]=2 \end{aligned} $

Let the sides of $\triangle A B C$ are $a-d, a$, $a+d$.

Given, perimeter $=42$

$ \begin{gathered} a-d+a+a+d=42 \Rightarrow a=14 \\ \text { As, } \quad B < A < C \\ A C < B C < A B \\ a-d < a < a+d \\ \Rightarrow \quad B C=14 \text { units } \\ \quad 2 R=\frac{a}{\sin A} \Rightarrow 2 \times \frac{65}{8}=\frac{B C}{\sin A} \\ \sin A=\frac{14 \times 8}{2 \times 65}=\frac{56}{65} \end{gathered} $

$ \text { } \begin{aligned}Let \,\, z= & b \tan \frac{B}{2} \tan \frac{C-A}{2} \\ = & b \tan \left[90-\left(\frac{C+A}{2}\right)\right] \tan \left(\frac{C-A}{2}\right) \\ = & b \cot \left(\frac{C+A}{2}\right) \tan \left(\frac{C-A}{2}\right) \\ = & b\left[\frac{2 \cos \left(\frac{C+A}{2}\right) \sin \left(\frac{C-A}{2}\right)}{2 \sin \left(\frac{C+A}{2}\right) \cos \left(\frac{C-A}{2}\right)}\right] \\ = & b\left[\frac{\sin C-\sin A}{\sin C+\sin A}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow \quad \frac{17}{22}=\frac{b^2+121-49}{2 \times 11 \times b} \\ & \Rightarrow \quad 17 b=b^2+72 \\ & \Rightarrow \quad b^2-17 b+72=0 \\ & \Rightarrow \quad b=9 \text { or } 8 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \sin A=\sqrt{1-\cos ^2 A}=\sqrt{1-\left(\frac{17}{22}\right)^2}=\sqrt{\frac{195}{22}} \\ & \sin C=\sqrt{1-\cos ^2 C}=\sqrt{1-\frac{1}{196}}=\frac{\sqrt{195}}{14} \end{aligned}\\ &\text { From Eq. (i), } \end{aligned} $

$ \begin{aligned} z & =b\left[\frac{\sin C-\sin A}{\sin C+\sin A}\right]=b\left[\frac{\frac{\sqrt{195}}{14}-\frac{\sqrt{195}}{22}}{\frac{\sqrt{195}}{14}+\frac{\sqrt{195}}{22}}\right] \\ & =b\left[\frac{22-14}{22+14}\right]=\frac{2 b}{9} \end{aligned} $

If $b=9$, then $z=2$

and if $b=8$, then $z=\frac{16}{9}$

$ \begin{aligned} & r^2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} \\ & =r^2\left[\frac{s(s-a)}{\Delta} \times \frac{s(s-b)}{\Delta} \times \frac{s(s-c)}{\Delta}\right] \\ & =\frac{r^2}{\Delta^3} \times s^2 \times s(s-a)(s-b)(s-c) \\ & =\frac{r^2 \times s^2}{\Delta^3} \times \Delta^2=\frac{r^2 \times s^2}{\Delta} \\ & =\frac{\Delta^2}{s^2} \times \frac{s^2}{\Delta} \,\,\,\,\left[\because r=\frac{\Delta}{s}\right]\\ & =\Delta \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} \sin A & =\frac{3 \sqrt{3}}{14}=\frac{\sqrt{27}}{14} \\ \cos A & =\sqrt{1-\sin ^2 A}\,\,\,\,\,\,\, \left[\because A<90^{\circ}\right]\\ & =\sqrt{1-\frac{27}{196}}=\frac{13}{14} \end{aligned} \end{aligned} $

By cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$

$ \begin{array}{ll} \Rightarrow & \frac{13}{14}=\frac{25+c^2-9}{2 \times 5 \times c} \\ \Rightarrow & 7 c^2-65 c+112=0 \\ \Rightarrow & 7 c^2-49 c-16 c+112=0 \\ \Rightarrow & \quad(c-7)(7 c-16)=0 \\ \Rightarrow & \quad c=7,16 / 7 \\ \therefore & \quad c=7 \end{array} $

[ $\because c$ is obtuse; $\therefore c$ is greatest]

$ \begin{aligned} &\begin{aligned} & (b \sin C+c \sin B)(b \cos C+c \cos B) \\ & =(b \sin C+c \sin B)(a) \end{aligned}\\ &\text { [By projection formula] }\\ &\begin{aligned} & =a b \sin C+a c \sin B \\ & =2 \Delta+2 \Delta \\ & {\left[\because \Delta=\frac{1}{2} a b \sin C \text { and } \Delta=\frac{1}{2} a c \sin B\right]} \\ & =4 \Delta \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \\ & \because \quad A=\pi r^2 \Rightarrow r=\sqrt{A / \pi} \\ & \therefore \quad r_1=\sqrt{A_1 / \pi} \\ & \quad r_2=\sqrt{\frac{A_2}{\pi}} \text { and } r_3=\sqrt{\frac{A_3}{\pi}} \\ & \Rightarrow \quad \frac{\sqrt{\pi}}{\sqrt{A}}=\frac{\sqrt{\pi}}{\sqrt{A_1}}+\frac{\sqrt{\pi}}{\sqrt{A_2}}+\frac{\sqrt{\pi}}{\sqrt{A_3}} \\ & \Rightarrow \quad \frac{1}{\sqrt{A}}=\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}} \\ & \Rightarrow \quad \frac{1}{\sqrt{A}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12} \Rightarrow A=\frac{144}{169} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { In } \triangle A B C, \\ & \begin{aligned} (b+c)^2 \sin ^2 \frac{A}{2}+(b-c)^2 \cos ^2 \frac{A}{2} & \\ = & K(1-\cos 2 A) \\ \Rightarrow(b+c)^2 \sin ^2 \frac{A}{2}+(b-c)^2( & \left.1-\sin ^2 \frac{A}{2}\right) \\ = & K\left(2 \sin ^2 A\right) \\ \Rightarrow(b-c)^2+\left\{(b+c)^2-(b-c)^2\right\} & \sin ^2 \frac{A}{2} \\ = & 2 K \sin ^2 A \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { On putting }\\ &\begin{aligned} & \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}, \sin A=\frac{a}{2 R} \\ & \text { So, }(b-c)^2+4 b c\left(\frac{(s-b)(s-c)}{b c}\right) \\ & =2 K\left(\frac{a^2}{4 R^2}\right) \\ & \Rightarrow(b-c)^2+4\left(\frac{a+b+c}{2}-b\right) \\ & \left(\frac{a+b+c}{2}-c\right)=\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+(a+c-b)(a+b-c)=\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+[a-(b-c)][a+(b-c)] \\ & =\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+\left[a^2-(b-c)^2\right]=\frac{K a^2}{2 R^2} \\ & \Rightarrow a^2=\frac{K a^2}{2 R^2} \Rightarrow K=2 R^2 \end{aligned} \end{aligned} $

In $\triangle A B C, b=7, c=4 \sqrt{3}$ and $A=\frac{\pi}{6}$

$ \begin{aligned} & \because \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow \cos \frac{\pi}{6}=\frac{49+48-a^2}{2 \times 7 \times 4 \sqrt{3}} \\ & \Rightarrow \frac{\sqrt{3}}{2}=\frac{97-a^2}{56 \sqrt{3}} \\ & \Rightarrow 84=97-a^2 \\ & a^2=13 \text { and } a=\sqrt{13} \end{aligned} $

Using sine rule,

$ \begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \frac{\sin \frac{\pi}{6}}{\sqrt{13}}=\frac{\sin B}{7}=\frac{\sin C}{4 \sqrt{3}} \\ & \sin B=\frac{7}{2 \sqrt{13}}, \sin C=\frac{2 \sqrt{3}}{\sqrt{13}} \end{aligned} $

$ \begin{aligned} \therefore a \sin B \sin C & =\sqrt{13} \times \frac{7}{2 \sqrt{13}} \times \frac{2 \sqrt{3}}{\sqrt{13}} \\ & =\frac{7 \sqrt{3}}{\sqrt{13}} \end{aligned} $

$ \begin{aligned} &\text { In } \triangle A B C \text {, }\\ &\begin{aligned} & r_1=\frac{\Delta}{s-a^{\prime}}, r_2=\frac{\Delta}{s-b} \text { and } r_3=\frac{\Delta}{s-c} \\ \because & r_2+r_3=\Delta\left(\frac{1}{s-b}+\frac{1}{s-c}\right) \\ = & \Delta\left(\frac{(s-c)+(s-b)}{(s-b)(s-c)}\right) \\ = & \Delta\left(\frac{2 s-b-c}{(s-b)(s-c)} \times \frac{s(s-a)}{s(s-a)}\right) \\ = & \Delta\left(\frac{a s(s-a)}{\Delta^2}\right) \\ & r_2+r_3=\frac{a s(s-a)}{\Delta}=\frac{a}{\tan A / 2}=a \\ \quad & \quad\left[\because A=\frac{\pi}{2}\right] \end{aligned} \end{aligned} $

Given, in a $\triangle A B C$,

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$

We know that,

$\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \cos C=\frac{a^2+b^2-c^2}{2 a b} \\ & \text { Since, } \frac{b^2+c^2-a^2}{2 a b c}+\frac{a^2+c^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2 a b c} \\ & =\frac{a^2+b^2+c^2}{2 a b c} . \end{aligned}$

Given,

$r_1=36, r_2=18 \text { and } r_3=12$

We know that, in a $\triangle A B C$

$\begin{aligned} \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \\ \Rightarrow \frac{1}{r} & =\frac{1}{36}+\frac{1}{18}+\frac{1}{12} \Rightarrow \frac{1}{r}=\frac{2+4+6}{72} \\ r & =6 \end{aligned}$

Now, we know that,

$\begin{aligned} \Delta^2 & =r \cdot r_1 \cdot r_2 \cdot r_3=6 \times 36 \times 18 \times 12 \\ \Delta^2 & =6^2 \times 6^2 \times 6^2 \\ \Delta & =6 \times 6 \times 6=216 \end{aligned}$

Again we know that,

$\begin{aligned} r & =\frac{\Delta}{S}=\frac{216}{S} \\ \Rightarrow \quad & S=\frac{216}{6}=36 \end{aligned}$

In a $\triangle A B C, a=6$ units, $b=5$ units and $c=4$ units

As we know that,

$\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c} \\ \Rightarrow \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{5^2+4^2-6^2}{2 \times 5 \times 4}=\frac{1}{8} \end{aligned}$

As we know that,

$\begin{aligned} \cos 2 A & =\cos ^2 A-\sin ^2 A \\ & =2 \cos ^2 A-1=2 \times\left(\frac{1}{8}\right)^2-1 \\ & =\frac{1}{32}-1=\frac{-31}{32} \end{aligned}$

Within the context of triangle $ \triangle ABC $, the following relationship holds:

We know that:

$ \left(\tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2}\right) \leq \frac{1}{3 \sqrt{3}} $

Thus, squaring both sides of the inequality, we get:

$ \begin{aligned} \left(\tan \frac{A}{2} \cdot \tan \frac{B}{2} \cdot \tan \frac{C}{2}\right)^2 & \leq \left(\frac{1}{3 \sqrt{3}}\right)^2 \\ & = \frac{1}{27} \end{aligned} $

Given that $A B C$ be triangle.

Using cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$

$\cos B=\frac{c^2+a^2-b^2}{2 c a}, \cos C=\frac{a^2+b^2-c^2}{2 a b}$

Now, $2(b c \cos A+c a \cos B+a b \cos C)$

$=2\binom{b c \frac{b^2+c^2-a^2}{2 b c}+c a \frac{c^2+a^2-b^2}{2 c a}}{+a b \cdot \frac{a^2+b^2-c^2}{2 a b}}$

$=2\left(\frac{b^2+c^2-a^2}{2}+\frac{c^2+a^2-b^2}{2}+\frac{a^2+b^2-c^2}{2}\right)$

$=2\left(\frac{b^2+c^2+a^2}{2}\right)=a^2+b^2+c^2$

Given that $A B C$ is a triangle

So, $A+B+C=180^{\circ}$

Given, $C=60^{\circ}$

$\Rightarrow \quad A+B=120^{\circ} \quad \text{... (i)}$

We have by sine formula

$\begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b} \\ \Rightarrow & \frac{\sin A}{\sin B}=\frac{a}{b}=2+\sqrt{3} \text { (given) } \end{aligned}$

Applying componendo and dividendo, we get

$\begin{aligned} & \frac{\sin A+\sin B}{\sin A-\sin B}=\frac{2-\sqrt{3}+1}{2+\sqrt{3}-1}=\frac{3+\sqrt{3}}{\sqrt{3}+1} \\ \Rightarrow & \frac{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)} \\ & =\frac{(3+\sqrt{3})(\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)} \\ \Rightarrow & \frac{2\left(\frac{\sqrt{3}}{2}\right)}{2\left(\frac{1}{2}\right)} \cot \left(\frac{A-E}{2}\right)=\sqrt{3} \\ \Rightarrow & \sqrt{3} \cot \left(\frac{A-B}{2}\right)=\sqrt{3} \Rightarrow\left(\frac{A-B}{2}\right)=45^{\circ} \\ \Rightarrow & A-B=90 \quad \text{.... (ii)} \end{aligned}$

Solving Eqs. (i) and (ii), we get $\angle A=105^{\circ}$

Given, $a=2, b=3, c=4$

Thus,

$\begin{aligned} \cos C & =\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{4+9-16}{2 \times 2 \times 3}=\frac{-3}{12}=\frac{-1}{4} \end{aligned}$

$\begin{aligned} & (b+c) \cos A+(c+a) \cos B+(a+b) \cos C \\ & =b \cos A+c \cos A+c \cos B+a \cos B+a \cos C+b \cos C \\ & =(b \cos C+c \cos B)+(c \cos A+a \cos C) \\ & \quad \quad+(a \cos B+b \cos A) \quad \ldots \text { (i) } \end{aligned}$

Using projection formula,

$\begin{aligned} & a=(b \cos C+c \cos B) \\ & b=(c \cos A+a \cos C) \\ & c=(a \cos B+b \cos A) \end{aligned}$

Substituting above values in Eq. (i), we get

$a+b+c$

Thus, $(b+c) \cos A+(c+a) \cos B+(a+b) \cos C =a+b+c$

Let the coordinates of C be (x, y).

Slope of $A C \times$ Slope of $B C=-1$

$\begin{aligned} & \left(\frac{y-3}{x-2}\right) \times\left(\frac{y-5}{x-4}\right)=-1 \\ \Rightarrow & y^2-8 y+15=-x^2+6 x-8 \\ \Rightarrow & x^2+y^2-6 x-8 y+23=0 \quad \text{... (i)}\\ & (A C)^2=(B C)^2 \text{[ABC is an isosceles right angled triangle]} \end{aligned}$

$\begin{aligned} \Rightarrow & (x-2)^2+(y-3)^2=(x-4)^2+(y-5)^2 \\ \Rightarrow & x^2+4-4 x+y^2+9-6 y \\ & =x^2+16-8 x+y^2+25-10 y \\ \Rightarrow & 4 x+4 y=28 \Rightarrow x+y=7 \\ \Rightarrow & y=7-x\quad \text{... (ii)} \end{aligned}$

Substituting the value of $y$ from Eq. (ii) in the Eq. (i),

$\begin{aligned} & x^2+(7-x)^2-6 x-8(7-x)+23=0 \\ \Rightarrow & x^2+49+x^2-14 x-6 x-56+8 x+23=0 \\ \Rightarrow & 2 x^2-12 x+16=0 \Rightarrow x^2-6 x+8=0 \\ \Rightarrow & x=4,2 \end{aligned}$

If $x=4$, then $y=3$ and If $x=2$, then $y=5$. If $(x, y)=(4,3)$, then

Centroid of triangle is $\left(\frac{2+4+4}{3}, \frac{3+3+5}{3}\right) =\left(\frac{10}{3}, \frac{11}{3}\right)$

If $(x, y)=(2,5)$, then centroid of $\triangle A B C$ is

$\left(\frac{2+2+4}{3}, \frac{5+3+5}{3}\right)=\left(\frac{8}{3}, \frac{13}{3}\right)$

$\begin{aligned} & \text { } \text { In } \triangle A B C, a \neq b \\ & \frac{a \cdot \cos A-b \cos B}{a \cos B-b \cos A}+\cos C \\ & =\frac{a\left[\frac{b^2+c^2-a^2}{2 b c}\right]-b\left[\frac{a^2+c^2-b^2}{2 a c}\right]}{a\left[\frac{a^2+c^2-b^2}{2 a c}\right]-b\left[\frac{b^2+c^2-a^2}{2 b c}\right]}+\cos C \\ & =\frac{a^2\left(b^2+c^2-a^2\right)-b^2\left(a^2+c^2-b^2\right)}{a b\left(a^2+c^2-b^2\right)-a b\left(b^2+c^2-a^2\right)}+\cos C \\ & =\frac{a^2 b^2+a^2 c^2-a^4-b^2 a^2-b^2 c^2+b^4}{a^3 b+a b c^2-a b^3-a b^3-a b c^2+a^3 b} \\ & +\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a^3 b-2 a b^3}+\frac{a^2+b^2-c^2}{2 a b} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a b\left(a^2-b^2\right)}+\frac{\left(a^2-b^2\right)\left(a^2+b^2-c^2\right)}{2 a b\left(a^2-b^2\right)} \\ & =\frac{a^2 c^2-a^4-b^2 c^2+b^4}{2 a b\left(a^2-b^2\right)} \\ \end{aligned}$

$+\frac{a^4+a^2 b^2-a^2 c^2-b^2 a^2-b^4+b^2 c^2}{2 a b\left(a^2-b^2\right)}$

$\begin{aligned} & a^2 c^2-a^4-b^2 c^2+b^4+a^4+a^2 b^2 \\ = & \frac{-a^2 c^2-b^2 a^2-b^4+b^2 c^2}{2 a b\left(a^2-b^2\right)} \\ = & \frac{0}{2 a b\left(a^2-b^2\right)}=0 \end{aligned}$

In a triangle $ \triangle ABC $, the given sides are $ a = 2 $, $ b = 3 $, and $ c = 4 $.

We start by calculating the semi-perimeter ($ s $) of the triangle:

$ s = \frac{a + b + c}{2} = \frac{2 + 3 + 4}{2} = \frac{9}{2} $

Next, we use the formula for $ \tan\left(\frac{A}{2}\right) $:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} $

Substitute the values:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{\left(\frac{9}{2} - 3\right)\left(\frac{9}{2} - 4\right)}{ \frac{9}{2}\left( \frac{9}{2} - 2 \right) }} = \sqrt{\frac{\left(\frac{9}{2} - \frac{6}{2}\right)\left(\frac{9}{2} - \frac{8}{2}\right)}{\frac{9}{2}\left(\frac{9}{2} - \frac{4}{2}\right)}} $

Simplify the fractions:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(9 - 6)(9 - 8)}{9(9 - 4)}} = \sqrt{\frac{3 \times 1}{9 \times 5}} $

Finally, we have:

$ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{1}{15}} $

The angles of $\triangle A B C$ are in ratio $1: 2: 3$.

Let $A=\theta, B=2 \theta, C=3 \theta$

$\begin{aligned} & \because \quad A+B+C=180^{\circ} \\ & \Rightarrow \quad \theta+2 \theta+3 \theta=180^{\circ} \\ & \Rightarrow \quad 6 \theta=180^{\circ} \\ & \Rightarrow \quad \theta=30^{\circ} \\ \end{aligned}$

$\begin{aligned} & \therefore A=30^{\circ}, B=60^{\circ} \text { and } C=90^{\circ} \\ & \because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \end{aligned}$

$\begin{aligned} & a=k \sin A, b=k \sin B \text { and } c=k \sin c \\ & \begin{aligned} a: b: c & =k \sin 30^{\circ}: k \sin 60^{\circ}: k \sin 90^{\circ} \\ & =\frac{1}{2}: \frac{\sqrt{3}}{2}: 1 \\ & =1: \sqrt{3}: 2 \end{aligned} \end{aligned}$

We know, $r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}$ and $r_3=\frac{\Delta}{s-c}$

Also, $\cot \left(\frac{A}{2}\right)=\frac{s(s-a)}{\Delta}, \cot \left(\frac{B}{2}\right)=\frac{s(s-b)}{\Delta}$

and $\cot \left(\frac{C}{2}\right)=\frac{s(s-c)}{\Delta}$

$\therefore r_1 \cot \left(\frac{A}{2}\right)+r_2 \cot \left(\frac{B}{2}\right)+r_3 \cot \left(\frac{C}{2}\right)$

$=\frac{\Delta}{(s-a)} \cdot \frac{s(s-a)}{\Delta}+\frac{\Delta}{(s-b)} \cdot \frac{s(s-b)}{\Delta}+\frac{\Delta}{(s-c)} \cdot \frac{s(s-c)}{\Delta}$

$\begin{aligned} & =s+s+s \\ & =3 s \end{aligned}$

$\begin{aligned} & \text { }(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\ & =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right)+2 a b\left(\sin ^2 \frac{C}{2}-\cos ^2 \frac{C}{2}\right) \\ & =a^2+b^2+2 a b\left(1-2 \cos ^2 \frac{C}{2}\right) \\ & =a^2+b^2+2 a b(-\cos C) \\ & =a^2+b^2-2 a b \cos C \\ & =a^2+b^2-2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right) \\ & =a^2+b^2-a^2-b^2+c^2=c^2 \\ & \therefore(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}=c^2\end{aligned}$

$r_1=\frac{\Delta}{s-a}, r_2 \frac{\Delta}{s-b}$ and $r_3=\frac{\Delta}{s-c}$ ...... (i)

Also,

$\begin{aligned} \Delta^2 & =r r_1 r_2 r_3 \quad .....\text{(ii)}\\ \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \quad .....\text{(iii)}\end{aligned}$

From Eq. (iii), we get

$\begin{aligned} & \frac{1}{r}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1 \\ \therefore \quad r & =1 \end{aligned}$

From Eq. (ii), we get $\Delta^2=(1)(2) (3) (6) =36$

$\Rightarrow \quad \Delta=6$

Also, $r=\frac{\Delta}{s}$

$\begin{aligned} & \Rightarrow \quad l=\frac{6}{s} \Rightarrow s=6 \\ & r_1=\frac{\Delta}{s-a} \\ & \Rightarrow s-a=\frac{\Delta}{r_1} \Rightarrow a=s-\frac{\Delta}{r_1} \\ & \Rightarrow \quad a=6-\frac{6}{2}=6-3 \\ & \Rightarrow \quad a=3 \Rightarrow r_2=\frac{\Delta}{s-b} \\ & \Rightarrow \quad b=s-\frac{\Delta}{r_2}=6-\frac{6}{3} \Rightarrow b=4 \\ & r_3=\frac{\Delta}{s-c} \\ & \Rightarrow \quad c=s-\frac{\Delta}{r_3}=6-\frac{6}{6}=6-1 \\ & \Rightarrow \quad c=5 \\ & \therefore(a, b, c)=(3,4,5) \\ & \end{aligned}$

$\begin{aligned} & a \tan A+b \tan B=(a+b) \tan \left(\frac{A+B}{2}\right) \\ & \Rightarrow a\left(\tan A-\tan \left(\frac{A+B}{2}\right)\right)=b\left(\tan \left(\frac{A+B}{2}\right)-\tan B\right) \\ & \Rightarrow a\left(\frac{\sin A}{\cos A}-\frac{\sin \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}\right)=b\left(\frac{\sin \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A+B}{2}\right)}-\frac{\sin B}{\cos B}\right) \\ & \Rightarrow a\left[\frac{\sin \left(A-\frac{(A+B)}{2}\right)}{\cos A \cos \left(\frac{A+B}{2}\right)}\right]=b\left[\frac{\sin \left(\frac{A+B}{2}-B\right)}{\cos B \cos \left(\frac{A+B}{2}\right)}\right] \\ & a\left[\frac{\sin \frac{(A-B)}{2}}{\cos A}\right]=b\left[\frac{\sin \frac{(A-B)}{2}}{\cos B}\right] \\ & \Rightarrow(k \sin A)\left[\frac{\sin \left(\frac{A-B}{2}\right)}{\cos A}\right]=(k \sin B)\left[\frac{\sin \left(\frac{A-B}{2}\right)}{\cos B}\right] \\ & \left\{\begin{array}{l}\operatorname{In} \triangle A B C \\ a=k \sin A \\ b=k \sin B\end{array}\right\} \\ & \Rightarrow \quad \tan A=\tan B \\ & \Rightarrow \quad \angle A=\angle B \Rightarrow A=B \\ & \end{aligned}$

Given : $AD=4;\angle DAB=\frac{\pi}{6};\angle ABE=\frac{\pi}{3}$

$\begin{aligned} A O=\frac{8}{3} \text { and } O D= & \frac{4}{3} \\\\ & {[\because O \text { divides } A D \text { in } 2: 1] } \end{aligned}$

Area of $\triangle A B C=2( \text{Area of} ~\triangle A B E)$

$\begin{aligned} & =2\left[\frac{3}{2}(\text { Area of } \triangle A O B)\right] \\\\ & =3 \times \text { area of } \triangle A O B \\\\ & =3 \times \frac{1}{2} \times B O \times A O \\\\ & =3 \cdot \frac{1}{2} \cdot \frac{8}{3 \sqrt{3}} \cdot \frac{8}{3} \\\\ & \qquad\left\{\because \tan \frac{\pi}{6}=\frac{B O}{A O} \Rightarrow B O=\frac{1}{\sqrt{3}} \times \frac{8}{3}\right\} \\\\ & =\frac{32}{3 \sqrt{3}} \end{aligned}$

$\begin{aligned} 2 \Delta^2\left(a^2+b^2+c^2\right) & =a^2 b^2 c^2 \\ a^2+b^2+c^2 & =\left(\frac{a b c}{\Delta}\right)^2 \cdot \frac{1}{2}=8 R^2 \end{aligned}$

$[\because a b c / \Delta=4 R$ where $R$-circumradius

$\begin{aligned} & \because \quad a=2 R \sin A, b=2 R \sin B, c=2 R \sin C] \\ & \Rightarrow 4 R^2\left(\sin ^2 A+\sin ^2 B+\sin ^2 C\right)=8 R^2 \\ & \Rightarrow \quad \sin ^2 A+\sin ^2 B+\sin ^2 C=2 \\ & \Rightarrow \quad(\cos 2 A+\cos 2 B+\cos 2 C)=-1 \\ & \Rightarrow \quad-1-4 \cos A \cos B \cos C=-1 \\ & \cos A \cos B \cos C=0 \\ & \end{aligned}$

So, any one among $A, B$ and $C$ has to be $\frac{\pi}{2}$.

So, right angled triangle.

$\because r=\frac{\Delta}{s}, r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}$,

$r_3=\frac{\Delta}{s-c} \text { and } 2 s=a+b+c$

Then, $r_1+r_2+r_3-r=\frac{\Delta}{s-a}+\frac{\Delta}{s-b}+\frac{\Delta}{s-c}-\frac{\Delta}{s}$

$\begin{aligned} & =\Delta\left(\frac{2 s-(a+b)}{(s-a)(s-b)}\right)+\Delta\left(\frac{c}{s(s-c)}\right) \\ & =\Delta \frac{(2 s-2 s+c)}{(s-a)(s-b)}+\Delta\left(\frac{c}{s(s-c)}\right) \\ & =\Delta\left[\frac{c s(s-c)+c(s-a)(s-b)}{s(s-a)(s-b)(s-c)}\right] \\ & \quad\left[\because \Delta^2=s(s-a)(s-b)(s-c)\right] \\ & =\frac{\Delta c}{\Delta^2}(s(s-c)+(s-a)(s-b)) \\ & \left.=\frac{c}{\Delta}\left(s^2-s c+s^2-(a+b) s+a b\right)\right) \end{aligned}$

$\begin{aligned} & =\frac{c}{\Delta}\left(2 s^2-(a+b+c) s+a b\right) \\ & =\frac{c}{\Delta}\left(2 s^2-2 s^2+a b\right)[\because 2 s=a+b+c] \\ & =\frac{a b c}{\Delta}=4\left(\frac{a b c}{4 \Delta}\right)=4 R\end{aligned}$

In $\Delta ABC,a=3,b=4,\sin A=3/4$

$\therefore$ $\cos A = \sqrt {1 - {{\sin }^2}A} = \sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}$

But $\cos A = {{{b^2} + {c^2} - {a^2}} \over {2bc}}$

$ \Rightarrow {{\sqrt 7 } \over 4} = {{{4^2} + {c^2} - {3^3}} \over {2\,.\,4\,.\,c}}$

$\begin{array}{lrlrl}\Rightarrow 2 \sqrt{7} c =7+c^2 \\ \Rightarrow c^2-2 \sqrt{7} c+7 =0 \\ \Rightarrow (c-\sqrt{7})^2 =0 \\ \therefore c =\sqrt{7} \\ \text { Now, } \cos B =\frac{c^2+a^2-b^2}{2 a c} \\ \Rightarrow \cos B =\frac{7+9-16}{2 \cdot 3 \cdot \sqrt{7}}=0 \Rightarrow B=90 \Upsilon \\ \therefore \angle C B A =90 \Upsilon\end{array}$

$A=75 Y, B=45 \Upsilon$, then $C=180 Y-(75 Y+45 Y)=60 \Upsilon$

$\because \quad \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}=2 R$ where $R$ is circumradius

$\Rightarrow \quad a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$

Then, $b+c \sqrt{2}=2 R(\sin B+\sqrt{2} \sin C)$

$\begin{aligned} & =2 R(\sin 45 Y+\sqrt{2} \sin 60 Y) \\ & =2 R\left(\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}\right)=2 R\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right) \\ & =2 \cdot 2 R\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \\ & =2 \cdot(2 R \sin 75 Y) \\ & =2(2 R \sin A) \\ & =2 a \end{aligned}$

In $\triangle A B C$

$\begin{aligned} & a=2 R \sin A=4 R \sin \frac{A}{2} \cos \frac{A}{2} \\ & b=2 R \sin B=4 R \sin \frac{B}{2} \cos \frac{B}{2} \\ & r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ & r_1=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ & r_2=4 R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} \\ & r_3=4 R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \end{aligned}$

$\text { Now, } \begin{aligned} & a b-r_1 r_2=16 R^2 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{A}{2} \\ & \cos \frac{B}{2}\left(1-\cos ^2 \frac{C}{2}\right) \\ &= 16 R^2 \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{B}{2} \sin ^2 \frac{C}{2} \end{aligned}$

$\therefore \quad \frac{a b-r_1 r_2}{r_3}=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=r$

Let $A, B, C$ is represented as $\mathbf{a}, \mathbf{b}, \mathbf{c}$ respectively.

Mid-points $D, E, F$ is $\frac{\mathbf{a}+\mathbf{b}}{2}, \frac{\mathbf{a}+\mathbf{c}}{2}, \frac{\mathbf{b}+\mathbf{c}}{2}$ respectively.

$\begin{aligned} \text { Then, } \mathbf{B E} & =\frac{\mathbf{a}-2 \mathbf{b}+\mathbf{c}}{2}, \mathbf{A F}=\frac{\mathbf{b}-2 \mathbf{a}+\mathbf{c}}{2}, \\ \mathbf{C D} & =\frac{\mathbf{a}+\mathbf{b}-2 \mathbf{c}}{2} \\ \mathbf{B E}+\mathbf{A F} & =\frac{2 \mathbf{c}-\mathbf{a}-\mathbf{b}}{2}=\frac{-(\mathbf{a}+\mathbf{b}-2 \mathbf{c})}{2} \\ & =-\mathbf{C D}=\mathbf{D C} \end{aligned}$

Given that,

$ \begin{equation*} \frac{a^3+b^3+c^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Using sine rule for a triangle $A B C$

$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R $

{where $2 R \rightarrow$ circumradius }

$\therefore a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$

Substituting the values of $a, b, c$ in Eq

(i), we get

$\frac{(2 R \sin A)^3+(2 R \sin B)^3+(2 R \sin C)^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$

or $\frac{(2 R)^3\left(\sin ^3 A+\sin ^3 B+\sin ^3 C\right)}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$

or $(2 R)^3=8 \quad$ or $\quad 2 R=2$

For a triangle $A B C$, it is given that

$ c=9, s=10 $

$ \begin{aligned} \text { and } \quad \Delta & =10 \sqrt{2} \\ \therefore \tan \frac{A-B}{2} & =\frac{a-b}{a+b} \cot \frac{c}{2}=\frac{a-b}{a+b} \frac{s(s-c)}{\Delta} \\ & =\frac{a-b}{a+b} \times \frac{10 \times 1}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} \frac{a-b}{a+b} \end{aligned} $

So, $b\left[1+\sqrt{2} \tan \frac{A-B}{2}\right]$

$ =b\left[1+\left(\sqrt{2} \times \frac{1}{\sqrt{2}} \frac{a-b}{a+b}\right)\right] $

$ \begin{aligned} & =b\left[\frac{a+b+a-b}{a+b}\right] \\ & =b\left[\frac{2 a}{a+b}\right]=a\left[\frac{2 b}{a+b}\right]=a\left[\frac{(a+b)-(a-b)}{a+b}\right] \\ & =a\left[1-\frac{a-b}{a+b}\right]=a\left[1-\sqrt{2} \tan \frac{A-B}{2}\right] \end{aligned} $

Let a $\triangle A B C$ of sides $a, b, c$, having area △

Area $=\Delta$

$ \frac{1}{2} b c \sin A=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

and $\quad \frac{1}{2} a c \sin B=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $\quad \frac{1}{2} a b \sin c=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using cosine rule

$ a^2=b^2+c^2-2 b c \cos A $

and    $ b^2=a^2+c^2-2 a c \cos B $

and     $ c^2=a^2+b^2-2 a b \cos C $

On adding, we get

$ \begin{align*} & \quad a^2+b^2+c^2=2 a^2+2 b^2+2 c^2 \\ & \quad-2 a b \cos C-2 a c \cos B-2 b c \cos A \\ & \text { or } \quad a^2+b^2+c^2=2 \\ & (a b \cos C+a c \cos B+b c \cos A) \quad \text { } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Now, from Eq. (i)

$ b c=\frac{2 \Delta}{\sin A} $

Eq. (ii), $\quad a c=\frac{2 \Delta}{\sin B}$

Eq. (iii), $ a b=\frac{2 \Delta}{\sin C} $

Putting these values in Eq. (iv), we get

$ \begin{aligned} & a^2+b^2+c^2=2 \\ & \left(\frac{2 \Delta}{\sin C} \cos C+\frac{2 \Delta}{\sin B} \cos B+\frac{2 \Delta}{\sin A} \cos A\right) \\ & a^2+b^2+c^2=4 \Delta(\cot C+\cot B+\cot A) \\ & \text { or } \cot C+\cot B+\cot A=\frac{a^2+b^2+c^2}{4 \Delta} \\ & \text { or } \cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4 \Delta} \end{aligned} $

Let three sides are $x-1, x, x+1$.

Given, largest angle is twice the smallest angle

$ \begin{align*} \therefore \cos \theta & =\frac{(x+1)^2+(x)^2-(x-1)^2}{2 x(x+1)} \\ & =\frac{x^2+2 x+1+x^2-x^2+2 x-1}{2 x(x+1)} \\ & =\frac{x^2+4 x}{2 x(x+1)}=\frac{x+4}{2(x+1)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

By Sine rule,

$ \begin{align*} & & \frac{x+1}{\sin 2 \theta} & =\frac{x-1}{\sin \theta} \\ \Rightarrow & & \frac{x+1}{2 \sin \theta \cos \theta} & =\frac{x-1}{\sin \theta} \\ \Rightarrow & & \cos \theta & =\frac{x+1}{2(x-1)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eqs. (i) and (ii),

$ \begin{aligned} & & \frac{x+1}{2(x-1)} & =\frac{x+4}{2(x+1)} \\ \Rightarrow & & x & =5 \end{aligned} $

∴ Sides are 4, 5, 6

$ \begin{aligned} \Delta & =\frac{1}{2} \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} \\ & =\frac{1}{4} \sqrt{15 \times 7 \times 5 \times 3}=\frac{15}{4} \sqrt{7} \end{aligned} $

In $\triangle A B C, A D$ and $B E$ are medians of $\triangle A B C$

$ \begin{array}{rlrl} A D & =\frac{7}{2} \Rightarrow \angle D A B=\frac{\pi}{8} \\ \angle A B E & =\frac{\pi}{4} \\ \Rightarrow & A O =\frac{2}{3} A D=\frac{2}{3} \times \frac{7}{2}=\frac{7}{3} \end{array} $

$ \angle A O B=\pi-\left(\frac{\pi}{8}+\frac{\pi}{4}\right)=\left(\pi-\frac{3 \pi}{8}\right) $

By Sine rule, $\frac{A O}{\sin \frac{\pi}{4}}=\frac{B O}{\sin \frac{\pi}{8}}$

$ \begin{aligned} B O=\frac{\frac{7}{3} \times \sin \frac{\pi}{8}}{\sin \frac{\pi}{4}} & =\frac{7}{3} \times \frac{\sin \frac{\pi}{8}}{2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}} \\ & =\frac{7}{6 \cos \frac{\pi}{8}} \end{aligned} $

Area of $\triangle A O B=\frac{1}{2} A O \times B O \times \sin \angle A O B$

$ \begin{aligned} & =\frac{1}{2} \times \frac{7}{3} \times \frac{7}{6} \times \frac{\sin \left(\pi-\frac{3 \pi}{8}\right)}{\cos \frac{\pi}{8}} \\ & =\frac{7}{36} \times \frac{\sin 3 \frac{\pi}{8}}{\cos \frac{\pi}{8}}=\frac{7}{36} \\ & \quad\left[\because \sin \frac{3 \pi}{8}=\cos \frac{3 \pi}{8}\right] \end{aligned} $

Area of $\triangle A B C=3 \times \frac{7}{36}=\frac{7}{12}$

Given, sides of $\triangle A B C$ is $5 k, 5 k$ and $6 k$ respectively,

$ \begin{aligned} &A B C \text { is an isosceles triangle, }\\ &\begin{aligned} & \Rightarrow \quad A D=\sqrt{25 k^2-9 k^2}=4 k \\ & \Rightarrow \quad \tan \theta=\frac{3}{4} \\ &\,\,\,\, \quad \angle B A C=2 \theta \\ & \Rightarrow \quad \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\ & \Rightarrow \quad \tan 2 \theta=\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7} \\ & \Rightarrow \quad 2 \theta=\tan ^{-1}\left(\frac{24}{7}\right) \\ & \therefore \text { Largest angle is } \tan ^{-1}\left(\frac{24}{7}\right) . \end{aligned} \end{aligned} $

In $\triangle A B C$,

$ \left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B $

We know that,

$ \begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R \\ & \begin{array}{r} \because 4 R^2\left[\left(\sin ^2 B-\sin ^2 C\right) \cot A\right. \\ \left.\quad+\left(\sin ^2 C-\sin ^2 A\right) \cot B\right] \end{array} \\ & 4 R^2[\sin (B+C) \sin (B-C)] \\ & \left.\frac{\cos A}{\sin A}+\sin (C+A) \sin (C-A) \frac{\cos B}{\sin B}\right] \\ & 4 R^2\left[\frac{\sin A \sin (B-C) \cos A}{\sin A}\right. \left.+\frac{\sin B \sin (C-A) \cos B}{\sin B}\right] \\ & 4 R^2[-\sin (B-C) \cos (B+C)-\sin (C-A) \cos (C+A)] \\ & 2 R^2[-\sin 2 B+\sin 2 C+\sin 2 A-\sin 2 C] \\ & 2 R^2[\sin 2 A-\sin 2 B] \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\right) \\ & =\frac{\Delta^2}{a^2+b^2+c^2} \\ & {\left[\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2}\right]} \\ & =\frac{1}{a^2+b^2+c^2} \\ & {\left[(s-a)^2+(s-b)^2+(s-c)^2+s^2\right]} \\ & =\frac{1}{a^2+b^2+c^2} \end{aligned} \end{aligned} $

$ \begin{aligned} & {\left[\left(\frac{a+b+c}{2}-a\right)^2+\left(\frac{a+b+c}{2}-b\right)^2\right.} \\ & \left.\quad+\left(\frac{a+b+c}{2}-c\right)^2+\left(\frac{a+b+c}{2}\right)^2\right] \\ & =\frac{1}{a^2+b^2+c^2} \\ & {\left[\left(\frac{b+c-a}{2}\right)^2+\left(\frac{a+c-b}{2}\right)^2\right.} \\ & \left.\quad+\left(\frac{a+b-c}{2}\right)^2+\left(\frac{a+b+c}{2}\right)^2\right] \end{aligned} $

$ =\frac{1}{a^2+b^2+c^2} $

$ =\frac{1}{a^2+b^2+c^2}\left[\frac{4\left(a^2+b^2+c^2\right)}{4}\right]=1 $

We have,

$ \begin{aligned} & R: r_1: r: 5: 12: 2 \\ & \therefore R=5 k, r_1=12 k, \\ & r=2 k \end{aligned} $

If $\triangle A B C$ is right angle of $A$

and $a=10 k, b=6 k$,

$ c=8 k $

Then $R=5 k, r_1=12 k$

Possible

$ \begin{array}{ll} \therefore & a^2=b^2+c^2 \\ \therefore & r+r_3+r_2-r_1 \end{array} $

$ \begin{aligned} & \frac{\Delta}{s}+\frac{\Delta}{s-c}+\frac{\Delta}{s-b}+\frac{\Delta}{s-a} \\ & \left(\frac{\Delta}{s}-\frac{\Delta}{s-a}\right)+\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-b}\right) \\ & \Rightarrow \Delta\left[\frac{s-a-s}{s(s-a)}+\frac{s-b+s-c}{(s-c)(s-b)}\right) \\ & \Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{2 s-b-c}{(s-c)(s-b)}\right] \\ & \Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{a}{(s-c)(s-b)}\right] \\ & \Rightarrow \Delta a\left[\frac{-(s-c)(s-b)+s(s-a)}{s(s-a)(s-b)(s-c)}\right] \\ & \Rightarrow \Delta a \frac{\left(s^2-a s-s^2+(b+c) s-b c\right)}{\Delta^2} \\ & \Rightarrow \frac{a^2-\left(b^2+c^2\right)}{\Delta^2}=0=\cos A \\ & \therefore \cos A \end{aligned} $

In $\triangle A B C$, we have

$ \begin{aligned} & \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{A C}{\sin B} \\ \Rightarrow \quad & \frac{A B}{\sin \left(180^{\circ}-4 \theta\right)}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \\ \Rightarrow \quad & \frac{A B}{\sin 4 \theta}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \end{aligned} $

Now, $\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta}$

$ \begin{aligned} & \Rightarrow \quad 16 \sin \theta=9 \sin 3 \theta \\ & \Rightarrow \quad 16 \sin \theta=9\left(3 \sin \theta-4 \sin ^3 \theta\right) \\ & \Rightarrow \quad 16=27-36 \sin ^2 \theta \Rightarrow \\ & 36 \sin ^2 \theta=11 \\ & \Rightarrow \quad \sin \theta=\frac{\sqrt{11}}{6} \Rightarrow \cos \theta=\frac{5}{6} \end{aligned} $

Now, $\sin 4 \theta=4 \sin \theta \cos \theta\left(1-2 \sin ^2 \theta\right)$

$ =4 \times \frac{\sqrt{11}}{6} \times \frac{5}{6}\left(1-2 \times \frac{11}{36}\right)=\frac{35 \sqrt{11}}{162} $

Again, we have

$ \begin{aligned} \frac{A B}{\sin 4 \theta} & =\frac{9}{\sin \theta} \\ \Rightarrow \quad A B & =\frac{9 \sin 4 \theta}{\sin \theta}=9 \times \frac{35 \sqrt{11}}{162} \times \frac{6}{\sqrt{11}} \\ & =\frac{35}{3} \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} r_1+r_2=4 R\left(\sin \frac{A}{2} \cos \frac{B}{2}\right. & \cos \frac{C}{2} \left.+\sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right] \\ & =4 R \cos \frac{C}{2} \sin \left(\frac{A+B}{2}\right) \\ & =4 R \cos \frac{C}{2} \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =4 R \cos \frac{C}{2} \cdot \cos \frac{C}{2} \\ & =2 R\left(2 \cos ^2 \frac{C}{2}\right)=2 R(1+\cos C) \\ & \therefore \quad \frac{1+\cos C}{r_1+r_2}=\frac{1}{2 R} \end{aligned}\\ &\text { Similarly, }\\ &\begin{aligned} & \frac{1+\cos A}{r_2+r_3}=\frac{1}{2 R}, \frac{1+\cos B}{r_1+r_3}=\frac{1}{2 R} \\ \therefore \quad & \frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3} +\frac{1+\cos B}{r_1+r_3}=\frac{3}{2 R} \end{aligned} \end{aligned} $

We have,

$ \begin{array}{r} \cos A \cos B+\sin A \sin B \sin C=1 \\ \Rightarrow \quad \cos A \cos B+\sin A \sin B-\sin A \sin B \\ +\sin A \sin B \sin C=1 \\ \Rightarrow \cos (A-B)-\sin A \sin B(1-\sin C)=1 \\ \Rightarrow 1-\cos (A-B)+\sin A \sin B(1-\sin C) =0 \\ \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}\right)+\sin A \sin B(1-\sin C)=0 \end{array} $

⇒ Now, $0 \leq \sin A, \sin B, \sin C \leq 1$

$ \therefore \quad \sin \left(\frac{A-B}{2}\right)=0 $

and $\sin A \sin B(1-\sin C)=0$

$ \Rightarrow A-B=0 \text { and } \sin C=1 $

$ \begin{array}{lc} & {[\because \sin A, \sin B \neq 0]} \\ \Rightarrow & A=B \text { and } C=90^{\circ} \\ \therefore & A=B=45^{\circ}, C=90^{\circ} \\ \text { Now, } & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow & \frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 90^{\circ}} \\ \Rightarrow & a: b: c=\frac{1}{\sqrt{2}}: \frac{1}{\sqrt{2}}: 1=1: 1: \sqrt{2} \end{array} $