Properties of Triangles

$b=3, c=\sqrt{19}$ and $\angle C=120^{\circ}$

$ \begin{array}{ll} \because & c^2=a^2+b^2-2 a b \cos C \\ \Rightarrow & 19=a^2+9-6 a \cdot \cos 120^{\circ} \\ \Rightarrow & 19=a^2+9+3 a \\ \Rightarrow & a^2+3 a-10=0 \\ \Rightarrow & a^2+5 a-2 a-10=0 \\ \Rightarrow & (a+5)(a-2)=0 \\ \because & a \neq-5 \end{array} $

Hence, $a=2$

$\because 2 A+C=300^{\circ}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $A+B+C=180^{\circ}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned}\Rightarrow & (2 A+C)-(A+B+C)=300^{\circ}-180^{\circ}=120^{\circ} \\ & \Rightarrow \quad A-B=120^{\circ} \Rightarrow A=B+120^{\circ} \end{aligned} $

By Eq. (i)

$ 2 B+C=60^{\circ} \Rightarrow C=60^{\circ}-2 B $

given that $R=8 r$

$ \begin{aligned} & \text { and } \frac{r}{R}=4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right) \\ & \Rightarrow \frac{1}{8}=4 \sin \left(\frac{A}{2}\right) \cdot \sin \left(\frac{B}{2}\right) \cdot \sin \left(\frac{C}{2}\right) \end{aligned} $

substitute values

$ \frac{1}{8 \times 4}=\sin \left(\frac{B}{2}+60^{\circ}\right) \sin \left(\frac{B}{2}\right) \sin \left(30^{\circ}+B\right) $

let $B / 2=u$

$ \begin{array}{r} \frac{1}{32}=\left(\frac{1}{2} \sin u+\frac{\sqrt{3}}{2} \cos u\right) \\ \quad \sin u\left(\frac{1}{2} \cos 2 u-\frac{\sqrt{3}}{2} \sin 2 u\right) \\ \Rightarrow \frac{1}{32}=\frac{1}{4} \sin u(\sin u+\sqrt{3} \cos u) \\ \quad(\cos 2 u-\sqrt{3} \sin 2 u) \end{array} $

$ \begin{aligned} \Rightarrow \frac{1}{32}=\frac{1}{4} \sin u \cdot\left(2 \sin \left(u+60^{\circ}\right)\right) .2 \cos \left(2 u+60^{\circ}\right) & \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{32}=\sin u \cdot \sin \left(u+60^{\circ}\right) \cdot \cos \left(2 u+60^{\circ}\right) \\ & \Rightarrow \frac{1}{32}=\left[\frac{1}{2}\left(\cos \left(u-u-60^{\circ}\right)-\cos \left(u+u+60^{\circ}\right)\right]\right. \cdot \cos \left(2 u+60^{\circ}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{32}=\left[\frac{1}{4}-\frac{1}{2} \cos \left(2 u+60^{\circ}\right)\right] \cdot \cos \left(2 u+60^{\circ}\right) \\ & \text { let } \cos \left(2 u+60^{\circ}\right)=t \\ & \therefore \quad \frac{1}{32}=\left(\frac{1}{4}-\frac{1}{2} t\right) t \\ & \Rightarrow \quad \frac{1}{32}=\frac{t}{4}-\frac{t^2}{2} \\ & \Rightarrow \quad \frac{1}{8}=t-2 t^2 \\ & \Rightarrow \quad 16 t^2-8 t+1=0 \\ & \therefore \quad t=\frac{1}{4} \end{aligned} $

$ \Rightarrow \cos \left(2 u+60^{\circ}\right)=\frac{1}{4} $

put $u=B / 2$

$ \Rightarrow \cos \left(B+60^{\circ}\right)=\frac{1}{4} $

Now, $C / 2=30^{\circ}-B$

and

$ \sin C / 2=\sin \left(30^{\circ}-B\right)=\cos \left(90^{\circ}-\left(130^{\circ}-B\right)\right) $

Hence, $\sin C / 2=1 / 4$

$ \begin{aligned} &\text { } \because \cos (A-B)=\cos A \cos B+\sin A \sin B\\ &\begin{aligned} & \text { and } \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \begin{aligned} & =\frac{16+c^2-25}{2 \times 4 \times c}=\frac{c^2-9}{8 c} \\ \cos B & =\frac{a^2+c^2-b^2}{2 a c}=\frac{c^2+9}{10 c} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { also, } \sin A=\sqrt{1-\cos ^2 A} \text { and } \\ & \begin{aligned} \sin B= & \sqrt{1-\cos ^2 B} \\ \Rightarrow \frac{31}{32} & =\left(\frac{c^2-9}{8 c}\right)\left(\frac{c^2+9}{10 c}\right) \\ & +\left(\sqrt{1-\left(\frac{c^2-9}{8 c}\right)^2}\right)\left(\sqrt{1-\left(\frac{c^2+9}{10 c}\right)^2}\right) \end{aligned} \end{aligned} $

from the options

$ \begin{aligned} & \text { put } C=6 \\ & \cos A=\frac{36-9}{8 \times 6}=9 / 16 \text { and } \\ & \cos B=\frac{36+9}{10 \times 6}=3 / 4 \\ & \begin{aligned} & \sin A=\sqrt{1-\left(\frac{9}{16}\right)^2}=\frac{\sqrt{175}}{16} \\ & \sin B=\sqrt{1-(3 / 4)^2}=\sqrt{7} / 4 \\ & \therefore \frac{9}{16} \times \frac{3}{4}+\frac{\sqrt{175}}{16} \times \frac{\sqrt{7}}{4} \\ &=\frac{27}{64}+\frac{35}{64}=31 / 32 \end{aligned} \end{aligned} $

Hence, $c=6$ satisfy the equation

Given, $A, B, C$ are in AP.

$ \begin{aligned} & \because A+B+C=180^{\circ} \\ & \Rightarrow 2 B=A+C \\ & \Rightarrow 2 B=180^{\circ}-B \Rightarrow B=60^{\circ} \\ & \cos B=\frac{1}{2} \end{aligned} $

Using cosine rule,

$ \begin{aligned} & \Rightarrow \frac{a^2+c^2-b^2}{2 a c}=\frac{1}{2} \\ & \Rightarrow a^2+c^2-b^2=a c \\ & \Rightarrow a^2+c^2-a c=b^2 \end{aligned} $

Let $A=60^{\circ}-\alpha, B=60^{\circ}, C=60^{\circ}+\alpha$

$ \begin{aligned} & \frac{a}{\sin \left(60^{\circ}-\alpha\right)}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin \left(60^{\circ}+\alpha\right)} \\ & =\frac{a+c}{\sin \left(60^{\circ}+\alpha\right)+\sin \left(60^{\circ}-\alpha\right)}=\frac{a+c}{\sqrt{3} \cos \alpha} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \frac{b}{\frac{\sqrt{3}}{2}}=\frac{a+c}{\sqrt{3} \cos \alpha} \\ & \Rightarrow 2 \cos \alpha=\frac{a+c}{b} \end{aligned}\\ &\text { Now, } \sqrt{a^2-a c+c^2} \cdot \cos \left(\frac{A-C}{2}\right)\\ &=\sqrt{b^2}(\cos \alpha)=b\left(\frac{a+c}{2 b}\right)=\frac{a+c}{2} \end{aligned} $

$ \begin{aligned} & \text {} \text { In } \triangle A B C, B=45^{\circ} \\ & a=2(\sqrt{3}+1) \end{aligned} $

$ \begin{aligned} & \text { Area }=6+2 \sqrt{3} \\ & \quad \frac{1}{2} a c \sin B=6+2 \sqrt{3} \\ & \begin{aligned} \frac{1}{2} \times 2(\sqrt{3}+1) c \sin 45^{\circ}= & 6+2 \sqrt{3} \\ & =2 \sqrt{3}(\sqrt{3}+1) \end{aligned} \\ & \Rightarrow \frac{c}{\sqrt{2}}=2 \sqrt{3} \Rightarrow c=2 \sqrt{6} \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \Rightarrow \frac{1}{\sqrt{2}}=\frac{4(4+2 \sqrt{3})+24-b^2}{8 \sqrt{6}(\sqrt{3}+1)} \\ & \Rightarrow 8 \sqrt{3}(\sqrt{3}+1)=40+8 \sqrt{3}-b^2 \\ & \Rightarrow b^2=40+8 \sqrt{3}-8 \sqrt{3}-24 \\ & \Rightarrow b^2=16 \Rightarrow b=4 \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} & \sin ^2 B=\sin A \\ & 2 \cos ^2 A=3 \cos ^2 B \\ \Rightarrow & 1-\cos ^2 B=\sin A \\ \Rightarrow & \cos ^2 B=\frac{2 \cos ^2 A}{3} \\ \Rightarrow & 1-\frac{2 \cos ^2 A}{3}=\sin A \\ \Rightarrow & 3-2 \cos ^2 A=3 \sin A \\ \Rightarrow & 1+2 \sin ^2 A=3 \sin A \\ \Rightarrow & 2 \sin A+\sin A-1=0 \\ & (2 \sin A-1)(\sin A-1)=0 \\ & \sin A=\frac{1}{2}, 1 \\ \therefore \quad & A=\frac{\pi}{6}, \frac{\pi}{2} \end{aligned} \end{aligned} $

Now, $\cos ^2 B=\frac{2}{3} \times \cos ^2 \frac{\pi}{6}=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}$

$ \Rightarrow \quad \cos B=\frac{1}{\sqrt{2}} \Rightarrow B=\frac{\pi}{4} $

If $\cos ^2 B=\frac{2}{3} \cos ^2 \frac{\pi}{2}=0$ (rejected)

$ \begin{aligned} \therefore A+B+C & =\pi \\ C & =\pi-\frac{\pi}{6}-\frac{\pi}{4}=\frac{12 \pi-2 \pi-3 \pi}{12} \\ & =\frac{7 \pi}{12} \end{aligned} $

$\therefore \triangle A B C$ is obtuse angled.

Given,

$ \begin{aligned} & \quad A=30^{\circ} \\ & \frac{b}{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}=\frac{c}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)} \\ & \Rightarrow \frac{b}{c}=\frac{(\sqrt{3}+1)^2+2(\sqrt{2}-1)}{(\sqrt{3}+1)^2-2(\sqrt{2}-1)} \end{aligned} $

By compounds and dividendo

$ \frac{b+c}{b-c}=\frac{2(\sqrt{3}+1)^2}{4(\sqrt{2}-1)} \Rightarrow \frac{b+c}{b-c}=\frac{2+\sqrt{3}}{\sqrt{2}-1} $

By compounds and dividendo

$ \begin{aligned} & \frac{2 b}{2 c}=\frac{2+\sqrt{3}+\sqrt{2}-1}{2+\sqrt{3}-\sqrt{2}+1} \\ & \Rightarrow \frac{b}{c}=\frac{\sqrt{3}+\sqrt{2}+1}{\sqrt{3}-\sqrt{2}+3} \Rightarrow \frac{\sin B}{\sin C}=\frac{\sqrt{3}+\sqrt{2}+1}{\sqrt{3}-\sqrt{2}+3} \\ & \Rightarrow \frac{\sin C}{\sin B}=\frac{3+\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}+1} \\ & \Rightarrow \frac{\sin \left(150^{\circ}-B\right)}{\sin B}=\frac{3+\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}+1} \\ & \quad\left\{\because A=30^{\circ} \Rightarrow B+C=150^{\circ}\right\} \\ & \Rightarrow \frac{1}{2} \cdot \frac{\cos B}{\sin B}+\frac{\sqrt{3} \sin B}{2} \frac{3+\sqrt{3}-\sqrt{2}}{\sin B}=\frac{\sqrt{3}+\sqrt{2}+1}{2} \\ & \Rightarrow \frac{\cot B}{2}=\frac{3+\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}+1}-\frac{\sqrt{3}}{2} \\ & \Rightarrow \frac{\cot B}{2}=\frac{6+2 \sqrt{3}-2 \sqrt{2}-3-\sqrt{6}-\sqrt{3}}{2(\sqrt{3}+\sqrt{2}+1)} \\ & \cot B=\frac{3+\sqrt{3}-2 \sqrt{2}-\sqrt{6}}{\sqrt{3}+\sqrt{2}+1}<0 \\ & \because \cot B=\text { Negative } \\ & \therefore \quad B>90^{\circ} \end{aligned} $

$ \begin{aligned} &\text { O-circumcentre }\\ &\begin{aligned} I & =\text { Incentre } \\ \cos A & =\frac{r}{R} \end{aligned} \end{aligned} $

$ \begin{aligned} & \because \cos A+\cos B+\cos C \\ & \quad=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} $

$ \begin{aligned} & =1+4 \cdot \frac{r}{4 R}\left(\because r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) \\ & =1+\frac{r}{R} \\ \therefore & \frac{r}{R}+\cos B+\cos C=1+\frac{r}{R} \\ \Rightarrow & \cos B+\cos C=1 \end{aligned} $

Let's first find the common ratio $r_1:r_2:r_3$.

We are given:

1. $r_1:r_2 = 3:4$

2. $r_2:r_3 = 2:3$

To combine these ratios, we need to make the value corresponding to $r_2$ common.

We can write the second ratio as:

$r_2:r_3 = 2:3 = (2 \times 2) : (3 \times 2) = 4:6$.

Now, we have:

$r_1:r_2 = 3:4$

$r_2:r_3 = 4:6$

Combining them gives the ratio $r_1:r_2:r_3 = 3:4:6$.

Next, we use the formulas for the exradii $r_1, r_2, r_3$ in terms of the area of the triangle $\Delta$ and the semi-perimeter $s = \frac{a+b+c}{2}$:

$r_1 = \frac{\Delta}{s-a}$

$r_2 = \frac{\Delta}{s-b}$

$r_3 = \frac{\Delta}{s-c}$

So, the ratio $r_1:r_2:r_3$ can be written as:

$\frac{\Delta}{s-a} : \frac{\Delta}{s-b} : \frac{\Delta}{s-c} = 3:4:6$.

Dividing by $\Delta$ (since $\Delta \neq 0$ for a triangle), we get:

$\frac{1}{s-a} : \frac{1}{s-b} : \frac{1}{s-c} = 3:4:6$.

This implies that there exists some constant $k$ (let's use $K$ to avoid confusion with $k$ for side ratios) such that:

$\frac{1}{s-a} = 3K \implies s-a = \frac{1}{3K}$

$\frac{1}{s-b} = 4K \implies s-b = \frac{1}{4K}$

$\frac{1}{s-c} = 6K \implies s-c = \frac{1}{6K}$

Let's denote $1/K$ as $C$ for simplicity. Then:

$s-a = C/3$

$s-b = C/4$

$s-c = C/6$

We know the relationships between the sides $a, b, c$ and $s-a, s-b, s-c$:

$a = (s-b) + (s-c)$

$b = (s-a) + (s-c)$

$c = (s-a) + (s-b)$

Substituting the expressions for $s-a, s-b, s-c$:

$a = \frac{C}{4} + \frac{C}{6} = \frac{3C+2C}{12} = \frac{5C}{12}$

$b = \frac{C}{3} + \frac{C}{6} = \frac{2C+C}{6} = \frac{3C}{6} = \frac{C}{2}$

$c = \frac{C}{3} + \frac{C}{4} = \frac{4C+3C}{12} = \frac{7C}{12}$

Now we can find the ratio $a:b:c$:

$a:b:c = \frac{5C}{12} : \frac{C}{2} : \frac{7C}{12}$

To simplify the ratio, we can multiply all parts by $12/C$:

$a:b:c = \left(\frac{5C}{12} \times \frac{12}{C}\right) : \left(\frac{C}{2} \times \frac{12}{C}\right) : \left(\frac{7C}{12} \times \frac{12}{C}\right)$

$a:b:c = 5 : 6 : 7$

The final answer is $\boxed{5:6:7}$.

Since, $a, b$ and $c$ are in AP

So, $2 b=a+c$

Also, $\angle A=2 \angle C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

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

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

So, $2 b=a+c$

$ \begin{array}{ll} \Rightarrow & 2 \cdot 2 R \sin B=2 R \sin A+2 R \sin C \\ \Rightarrow & 2 \sin B=\sin A+\sin C \end{array} $

In a $\triangle A B C$,

$ \begin{aligned} & A+B+C=180^{\circ} \\ \Rightarrow & 2 C+B+C=180^{\circ} \\ \Rightarrow & B=180^{\circ}-3 C \end{aligned} $

$ \text { } \begin{aligned}So, \sin B & =\sin \left(180^{\circ}-3 C\right) \\ & =\sin (3 C) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} & \text { And } 2 \sin B=\sin A+\sin C \\ & \Rightarrow 2 \sin (3 C)=\sin (2 C)+\sin C \end{aligned} $

[using Eq. (i) and (ii)]

$ =2 \sin C \cos C+\sin C $

and $\sin (3 C)=3 \sin C-4 \sin ^3 C$

$ \text { } \begin{aligned}So,2 \sin (3 C) & =2\left(3-4 \sin ^2 C\right) \\ & =2 \cos C+1 \\ & =2\left(4 \cos ^2 C-1\right) \end{aligned} $

$ \begin{aligned} \Rightarrow & 8 \cos ^2 C-2=2 \cos C+1 \\ \Rightarrow & 8 \cos ^2 C-2 \cos C-3=0 \\ \therefore & \cos C=\frac{2 \pm \sqrt{4-4(8)(-3)}}{16} \\ & =\frac{2 \pm \sqrt{100}}{16}=\frac{2 \pm 10}{16} \end{aligned} $

So, $\quad \cos C=\frac{3}{4}, \cos C=-\frac{1}{2}$

Since, $C$ must be positive,

So, $\cos C=\frac{3}{4}$

Now, $\cos A=\cos (2 C)$

$ =2 \cos ^2 C-1=2\left(\frac{3}{4}\right)^2-1=\frac{1}{8} $

And $B=180^{\circ}-3 C$

$ \begin{aligned} \cos B & =\cos \left(180^{\circ}-3 C\right) \\ & =-\cos (3 C) \\ & =-\left(4 \cos ^3 C-3 \cos C\right) \\ & =-\left(4\left(\frac{27}{64}\right)-3 \cdot \frac{3}{4}\right) \\ & =-\left(\frac{27}{16}-\frac{9}{4}\right) \\ & =-\left(\frac{27-36}{16}\right)=\frac{9}{16} \end{aligned} $

$ \therefore \cos B=\frac{9}{16} $

So, $\cos A: \cos B: \cos C$

$ \begin{aligned} & =\frac{1}{8}: \frac{9}{16}: \frac{3}{4} \\ & =\frac{2: 9: 12}{16} \end{aligned} $

So, $\cos A: \cos B: \cos C=2: 9: 12$

In a triangle $A, B$ and $C$ are in arithmetic progression.

So, $B=\frac{\pi}{3}$

Now, $r r_3=r_1 r_2$

$ \begin{aligned} & \Rightarrow \quad \frac{\Delta}{s} \cdot \frac{\Delta}{s-c}=\frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \\ & \Rightarrow \quad \frac{1}{s(s-c)}=\frac{1}{(s-a)(s-b)} \\ & \Rightarrow \quad s^2-(a+b) s+a b=s^2-c s \\ & \Rightarrow \quad a b=(a+b-c) s \\ & \Rightarrow \quad a b=(a+b-c) \frac{(a+b+c)}{2} \\ & \Rightarrow \quad 2 a b=(a+b)^2-c^2 \\ & \Rightarrow \quad 2 a b=a^2+b^2+2 a b-c^2 \\ & \Rightarrow \quad a^2+b^2=c^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, using cosine rule,

$ \begin{aligned} & b^2=a^2+c^2-2 a c \cos B \\ \Rightarrow & b^2=a^2+c^2-2 a c \cos \left(\frac{\pi}{3}\right) \\ \Rightarrow & b^2=a^2+c^2-a c \\ \Rightarrow & c^2-a^2=a^2+c^2-a c \\ \Rightarrow & 2 a^2=a c \Rightarrow 2 a=c \\ \Rightarrow & a=\frac{c}{2}=\frac{10}{2}=5 \end{aligned} $

So, $b^2=c^2-a^2=100-25=75$

$ \Rightarrow \quad b=5 \sqrt{3} $

Now, $a^2+b^2+c^2=25+75+100=200$

LHS $=\frac{2\left(r_1+r_3\right)}{a c(1+\cos B)}$

$ \begin{aligned} & =\frac{2\left(r_1+r_3\right)}{a c\left(1+2 \cos ^2 \frac{B}{2}-1\right)} \\ & =\frac{2\left(r_1+r_3\right)}{a c\left(2 \cos ^2 \frac{B}{2}\right)} \end{aligned} $

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

And $\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{a c}}$

Hence, LHS $=\frac{2\left[\frac{\Delta}{s-a}+\frac{\Delta}{s-c}\right]}{a c \cdot 2\left(\sqrt{\frac{s(s-b)}{a c}}\right)^2}$

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

$ \begin{aligned} & =\frac{\Delta(a+b+c-a-c)}{s(s-a)(s-b)(s-c)} \\ & =\frac{b \Delta}{s(s-a)(s-b)(s-c)} \\ & =\frac{b}{\Delta}=\text { RHS } \end{aligned} $

$\because a=8, b=10$ and $c=12$

and $r=\frac{\Delta}{s}, R=\frac{a b c}{4 \Delta}$

So, $\frac{r}{R}=\frac{4 \Delta^2}{s \cdot a b c}$

$ \because \quad s=\frac{a+b+c}{2}=15 $

also, $\Delta=\sqrt{15(15-8)(15-10)(15-12)}$

$ =\sqrt{15 \times 7 \times 5 \times 3}=15 \sqrt{7} $

Thus, $\frac{r}{R}=\frac{4 \times 225 \times 7}{15 \times 8 \times 10 \times 12}=\frac{7}{16}$

In $\triangle A B C, \angle A+\angle B+\angle C=180^{\circ}$

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

$ \begin{aligned} \Rightarrow \cos (B+C) & =\cos \left(180^{\circ}-A\right) \\ & =-\cos A \end{aligned} $

$ \begin{aligned}Now,\,\, \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{8^2+7^2-13^2}{2 \times 8 \times 7} \\ & =\frac{64+49-169}{112}=-\frac{56}{112} \\ & =-\frac{1}{2} \end{aligned} $

Hence, $\cos (B+C)=-\left(-\frac{1}{2}\right)=\frac{1}{2}$

$\because\left(r_1-r_3\right)\left(r_1-r_2\right)-2 r_2 r_3=0$

and we know that

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

substitute these values

$ \begin{array}{r} \Rightarrow \quad \Delta^2\left(\frac{1}{s-a}-\frac{1}{s-c}\right)\left(\frac{1}{s-a}-\frac{1}{s-b}\right) -2 \Delta^2\left(\frac{1}{(s-b)(s-c)}\right)=0 \\ \Rightarrow \quad\left(\frac{s-c-s+a}{(s-a)(s-c)}\right)\left(\frac{s-b-s+a}{(s-a)(s-b)}\right) -\left(\frac{2}{(s-b)(s-c)}\right)=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{1}{(s-b)(s-c)}\left[\frac{(a-c)(a-b)}{(s-a)^2}-2\right]=0 \\ & \Rightarrow \quad(a-c)(a-b)-2(s-a)^2=0 \\ & \Rightarrow \quad(a-c)(a-b)-2\left(\frac{a+b+c}{2}-a\right)^2=0 \\ & \Rightarrow \quad 2(a-c)(a-b)-(b+c-a)^2=0 \\ & \Rightarrow \quad 2\left(a^2-a b-a c+b c\right)-\left(b^2+c^2+a^2\right)-2(b c-c a-a b)=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad a^2-b^2-c^2-2 a b-2 a c+2 b c -2 b c+2 a c+2 a b=0 \\ & \Rightarrow \quad a^2-b^2=c^2 \end{aligned} $

$\because D$ is the mid-point of $B C$ and $E$ is the mid-point of $A D$.

Let $B E$ and $A C$ intersects at $F$

So, $\frac{A F}{F C}=\frac{1}{2}$

and $A C=A F+F C=\frac{1}{3} A C+\frac{2}{3} A C$

Thus, $\frac{A F}{A C}=\frac{\frac{1}{3} A C}{A C}=\frac{1}{3}$

Hence, $A F: A C=1: 3$

We have,

$ \cos A \cos B+\sin A \cdot \sin B \sin C=1 $

In $\triangle A B C$,

$\cos A \cos B+\sin A \sin B \sin C=\mathrm{l}$ is possible if $A=45^{\circ}, B=45^{\circ}$ and $C=90^{\circ}$ i.e.

$ \begin{aligned} & \cos 45^{\circ} \cdot \cos 45^{\circ}+\sin 45^{\circ} \cdot \sin 45 \cdot \sin 90^{\circ} \\ & =\frac{1}{2}+\frac{1}{2}=1 \\ & \therefore \sin A+\sin B+\sin C \\ & \quad=\sin 45^{\circ}+\sin 45^{\circ}+\sin 90^{\circ} \\ & \quad=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1=\sqrt{2}+1 \end{aligned} $

Given, $a: b: c=4: 5: 6$

We can express the sides as $a=4 x$, $b=5 x$ and $c=6 x$ for some positive constant $x$.

$ \begin{aligned} & \because \cos A=\frac{b^2+c^2-a^2}{2 b c}(\text { By Cosine rule) } \\ & =\frac{(5 x)^2+(6 x)^2-(4 x)^2}{2 \times 5 x \times 6 x}=\frac{45 x^2}{60 x^2}=\frac{3}{4} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} Similarly,\,\, \cos B & =\frac{(4 x)^2+(6 x)^2-(5 x)^2}{2(4 x)(6 x)} \\ & =\frac{27 x^2}{48 x^2}=\frac{9}{16} \end{aligned} \\ & \text { and } \cos C=\frac{(4 x)^2+(5 x)^2-(6 x)^2}{2(4 x)(5 x)} \\ & =\frac{5 x^2}{40 x^2}=\frac{1}{8} \end{aligned} $

$ \text { } \begin{aligned}So, \,\, \frac{\cos A+3 \cos C}{\cos B} & =\frac{\frac{3}{4}+3\left(\frac{1}{8}\right)}{\frac{9}{16}}=\frac{\frac{6+3}{8}}{\frac{9}{16}} \\ & =\frac{9}{8} \times \frac{16}{9}=2 \end{aligned} $

Given, $a=6, b=8, c=10$

$ \begin{aligned} & \therefore \quad s=\frac{a+b+c}{2}=12 \\ & \text { and area }(\Delta)=\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{12 \times 6 \times 4 \times 2}=24 \end{aligned} $

Now, $r_1=\frac{\Delta}{s-a}=\frac{24}{6}=4$ and

$ \begin{aligned} r & =\frac{\Delta}{s}=\frac{24}{12}=2 \\ r_2 & =\frac{\Delta}{s-b}=\frac{24}{4}=6 \\ r_3 & =\frac{\Delta}{s-c}=\frac{24}{2}=12 \\ \therefore \quad \frac{2 r_2 r_3}{r_1} & =\frac{2 \times 6 \times 12}{2 \times 4}=18 \\ & =8+10=b+c \end{aligned} $

$a, b, c$ are in HP.

$ \Rightarrow b=\frac{2 a c}{a+c} $

Let $\operatorname{cosec}^2 \frac{A}{2}, \operatorname{cosec}^2 \frac{B}{2}, \operatorname{cosec}^2 \frac{C}{2}$ are in AP .

$ \begin{aligned} & \Rightarrow \frac{1}{\sin ^2 \frac{A}{2}}, \frac{1}{\sin ^2 \frac{B}{2}}, \frac{1}{\sin ^2 \frac{C}{2}} \text { are in AP. } \\ & \Rightarrow \frac{b c}{(s-b)(s-c)}, \frac{a c}{(s-a)(s-c)} \\ & \frac{a b}{(s-a)(s-b)} \text { are in AP. } \\ & \Rightarrow \frac{b c}{s(s-b)(s-c)}, \frac{a c}{s(s-a)(s-c)} \\ & \frac{a b}{s(s-a)(s-b)} \text { are in AP. } \end{aligned} $

$\Rightarrow\frac{b c(s-a)}{s(s-a)(s-b)(s-c)}, \frac{a c(s-b)}{s(s-a)(s-b)(s-c)^{\prime}} \frac{a b(s-c)}{s(s-a)(s-b)(s-c)}$ are in AP.

$\Rightarrow b c s-a b c, a c s-a b c, a b s-a b c$ are in AP.

$\Rightarrow b c s, a c s, a b s$ are in AP.

$\Rightarrow b c, a c, a b$ are in AP.

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP.

$\Rightarrow a, b, c$ are in HP. Which is true.

$ \text { } \begin{aligned} & r=3, R=5 \\ & \begin{aligned} \frac{1}{a b}+ & \frac{1}{b c}+\frac{1}{c a}=\frac{a+b+c}{a b c}=\frac{2 s}{a b c} \\ & =\frac{2 s}{a b c} \times \frac{\Delta}{\Delta} \\ & =\frac{1}{2} \frac{4 \Delta}{a b c} \cdot \frac{s}{\Delta} \\ & =\frac{1}{2} \cdot \frac{1}{R} \cdot \frac{1}{r}=\frac{1}{2} \cdot \frac{1}{5} \cdot \frac{1}{3}=\frac{1}{30} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given in } \triangle A B C \text {, }\\ &\begin{aligned} & A-B=120^{\circ}, R=8 r \\ & \frac{1+\cos C}{1-\cos C} \\ \Rightarrow & R=8\left(4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) \\ \Rightarrow & 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{1}{16} \\ \Rightarrow & \left(\cos \frac{A-B}{2}-\cos \left(\frac{A+B}{2}\right)\right) \sin \frac{C}{2}=\frac{1}{16} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & \sin \frac{C}{2}\left(\frac{1}{2}-\sin \frac{C}{2}\right)=\frac{1}{16} \\ & \sin \frac{C}{2}=\frac{1}{4} \\ & \cos C=1-\frac{2}{16}=\frac{7}{8} \\ \Rightarrow & \frac{1+\cos C}{1-\cos C}=15 \end{aligned} $

$ \begin{aligned} &\text { Given, } \sqrt{\frac{r_2}{r_3 r_1}} \text {, in } \triangle A B C\\ &\begin{aligned} \sqrt{\frac{r_2}{r_3 r_1}} & =\sqrt{\frac{\frac{\Delta}{s}-\frac{\Delta}{s-b}}{\frac{\Delta}{s-c} \frac{\Delta}{s-a}}} \sqrt{\frac{\frac{\Delta}{s} \frac{\Delta}{s-b}}{\frac{\Delta}{s-c} \frac{\Delta}{s-a}}} \\ & =\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \\ \Rightarrow s & =\frac{a+b+c}{2} \\ \Rightarrow 2 s & =a+b+c \Rightarrow 2 s-a-c=b \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} Now, & \frac{b}{r_3+r_1}=\frac{b}{\frac{\Delta}{s-c}+\frac{\Delta}{s-a}} \\ & =\frac{b}{\frac{\Delta(2 s-a-c)}{(s-a)(s-c)}}=\frac{b(s-a)(s-c)}{\Delta b} \\ & =\frac{(s-a)(s-c)}{\sqrt{s(s-a)(s-b)(s-c)}} \\ & =\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\frac{r r_2}{r_3 r_1}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } x \text {-coordinate of incentre }\\ &=\frac{a x_1+b y_1+c z_1}{a+b+c} \end{aligned} $

$ a=\sqrt{98} $

$ \begin{aligned} b & =13 \\ c & =5 \\ I_x & =\frac{\sqrt{98} \times 0+13 \times 3+5 \times 0}{18+\sqrt{98}} \\ & =\frac{39}{18+\sqrt{98}}=\frac{39}{18+7 \sqrt{2}} \end{aligned} $

$\sin \frac{A}{2}=\frac{1}{4} \sqrt{\frac{3}{5}} a=2 c=5$

$ \begin{aligned} & \cos A=1-2 \sin ^2 \frac{A}{2}=1-2 \cdot \frac{1}{16} \times \frac{3}{5}=\frac{37}{40} \\ & \text { Now, } \cos A=\frac{b^2+c^2-a^2}{2 \times b \times c} \\ & \Rightarrow \quad \frac{37}{40}=\frac{b^2+25-4}{2 \times b \times 5} \\ & \Rightarrow \quad \frac{37}{4}=\frac{b^2+21}{b} \\ & \Rightarrow 4 b^2-37 b+84=0 \\ & \therefore(4 b-21)(b-4)=0 \end{aligned} $

$\therefore b=4$ because given $b$ is an integer

$ \begin{aligned} & \because \sin A=\sqrt{1-\cos ^2 A}=\sqrt{1-\left(\frac{37}{40}\right)^2} \\ & \therefore \text { Area of triangle }=\frac{1}{2} b \times c \times \sin A \\ & \quad=\frac{1}{2} \times 4 \times 5 \sqrt{1-\left(\frac{37}{40}\right)^2} \\ & \quad=10 \sqrt{\frac{77}{40} \times \frac{3}{40}}=\frac{1}{4} \sqrt{231} \end{aligned} $

$ \begin{aligned} \\ &\begin{aligned} & \text { Given, } a+c=5 b \\ & a+b+c=2 s=6 b \\ & \begin{array}{l} s=3 b \\ \cot \frac{A}{2} \cdot \cot \frac{C}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \\ \quad \times \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \end{array} \\ & =\frac{s}{s-b}=\frac{3 b}{3 b-b}=\frac{3 b}{2 b}=\frac{3}{2} \end{aligned} \end{aligned} $

Given, $r_1=3, r_2=4$ and $r_3=6$

$ \begin{aligned} & r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c} \\ & \frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{s}{\Delta} \\ & \frac{1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{s}{\Delta} \Rightarrow \frac{s}{\Delta}=\frac{3}{4} \end{aligned} $

And $r_1 r_2+r_2 r_3+r_1 r_3=s^2$

$ \begin{array}{ll} \Rightarrow & 3 \times 4+4 \times 6+3 \times 6=s^2 \\ \Rightarrow & 12+24+18=s^2 \\ \Rightarrow & s^2=54 \\ \Rightarrow & s=3 \sqrt{6} \\ \therefore & \Delta=\frac{4 s}{3}=\frac{12 \sqrt{6}}{3}=4 \sqrt{6} \\ & r_2=\frac{\Delta}{s-b} \Rightarrow s-b=\frac{\Delta}{r_2} \\ & b=s-\frac{\Delta}{r_2}=3 \sqrt{6}-\frac{4 \sqrt{6}}{4} \\ & =3 \sqrt{6}-\sqrt{6}=2 \sqrt{6} \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & x=a+b \\ & y=a b \end{aligned} \end{aligned} $

We have, $x^2-c^2=y$

$ \begin{aligned} \Rightarrow & \quad(a+b)^2-c^2=a b \\ \Rightarrow & \quad a^2+b^2-c^2=-a b \\ \Rightarrow & \quad \frac{a^2+b^2-c^2}{2 a b}=-\frac{1}{2} \\ & \quad \cos c=-\frac{1}{2} \Rightarrow c=\frac{2 \pi}{3} \Rightarrow \sin c=\frac{\sqrt{3}}{2} \end{aligned} $

Using circumradius formula,

$ \begin{aligned} & c=2 R \sin c \\ & R=\frac{c}{\frac{2 \sqrt{3}}{2}}=\frac{c}{\sqrt{3}} \end{aligned} $

We have, $\tan \frac{B}{2}=\frac{\sqrt{5}}{4}$ and $b=6$

Area of $\triangle A B C=\triangle=4 \sqrt{5}$

$ \cos B=\frac{1-\tan ^2 \frac{B}{2}}{1+\tan ^2 \frac{B}{2}}=\frac{11}{21} $

$ \begin{aligned} \Rightarrow \quad \sin B & =\frac{\sqrt{320}}{21}=\frac{8 \sqrt{5}}{21} \\ \because \quad \Delta & =\frac{1}{2} a c \sin B=4 \sqrt{5} \\ & =\frac{1}{2} a c \frac{8 \sqrt{5}}{21}=4 \sqrt{5} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \because \quad a c=21 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \because \quad b^2=c^2+a^2-2 a c \cos B \\ & \Rightarrow \quad 36=c^2+a^2-2 \times 21 \times \frac{11}{21} \\ & \Rightarrow \quad 36=c^2+a^2-22 \\ & \Rightarrow \quad c^2+a^2=58 \\ & \because \quad(c+a)^2=c^2+a^2+2 a c \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad=58+42=100 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad c+a=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow \quad c=7 \text { and } a=3 \end{aligned}\\ &\because \text { Smallest sides is } 3 \text { units. } \end{aligned} $

$ \begin{aligned} &\text { We have, } r_1=2 r_2=3 r_3\\ &\Rightarrow \quad \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\frac{1}{k} \quad \text { (say) } \end{aligned} $

$ \begin{aligned}\because \,\,& s-a=k \\ & s-b=2 k \\ & s-c=3 k \\ & \hline \end{aligned} $

$ \begin{aligned} 3 s-(a+b+c) & =6 k \\ s & =6 k \\ a & =s-k=5 k \\ b & =s-2 k=4 k \\ \because \quad a: b & =5: 4 \end{aligned} $

In $\triangle ABC$, to evaluate the expression $a^2 \sin 2B + b^2 \sin 2A$, let's proceed with the following transformations:

$ \begin{aligned} &a^2 \sin 2B + b^2 \sin 2A \\ &= 2a^2 \sin B \cos B + 2b^2 \sin A \cos A \\ &= 2a(b \sin A) \cos B + 2b^2 \sin A \cos A \\ &= 2b \sin A (a \cos B + b \cos A). \end{aligned} $

Now, applying the sine rule and angle sum identity in $\triangle ABC$:

$ \begin{aligned} &a \cos B + b \cos A = c. \\ &\text{Thus, the expression simplifies to:} \\ &= 2b \sin A(c). \end{aligned} $

By the sine rule, we know that:

$ c = a \cos B + b \cos A \quad \Rightarrow \quad a^2 \sin 2B + b^2 \sin 2A = 2ab \sin C. $

Therefore, the expression simplifies to $2ab \sin C$.

$ \begin{aligned} &\text { Given, In } \triangle A B C \text {, }\\ &\begin{aligned} & \frac{r_2\left(r_1+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=\frac{r_1 r_2+r_2 r_3}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}} \\ & =\frac{\frac{\Delta}{(s-a)} \times \frac{\Delta}{(s-b)}+\frac{\Delta}{(s-b)} \times \frac{\Delta}{(s-c)}}{\sqrt{\frac{\Delta}{s-a} \times \frac{\Delta}{s-b}+\frac{\Delta}{s-b} \times \frac{\Delta}{s-c}}+\frac{\Delta}{s-c} \times \frac{\Delta}{s-a}} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{\frac{\Delta^2(s-c+s-a)}{(s-a)(s-b)(s-c)}}{\Delta \sqrt{\frac{(s-c)+(s-a)+(s-b)}{(s-a)(s-b)(s-c)}}} \\ & =\frac{\Delta(2 s-a-c)}{(s-a)(s-b)(s-c)} \times \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{3 c-a-b-c}} \\ & =\frac{\Delta(b)}{\sqrt{(s-a)(s-b)(s-c)} \sqrt{s}} \\ & =\frac{\Delta(b)}{\sqrt{s(s-a)(s-b)(s-c)}}=b \end{aligned} $

$ \begin{aligned} &\text { We have, in } \triangle A B C\\ &\begin{aligned} & \left(r_2+r_3\right) \operatorname{cosec}^2 \frac{A}{2} \\ & =\binom{4 R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}}{+4 R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}} \operatorname{cosec}^2 \frac{A}{2} \\ & =4 R \cos \frac{A}{2}\left(\sin \frac{B}{2} \cos \frac{C}{2}+\cos \frac{B}{2} \sin \frac{C}{2}\right) \\ & =4 R \cos \frac{A}{2} \sin \left(\frac{B}{2}+\frac{C}{2}\right) \operatorname{cosec}^2 \frac{A}{2} \\ & =4 R \cos \frac{A}{2} \cdot \cos \frac{A}{2} \operatorname{cosec}^2 \frac{A}{2} \\ & =4 R \cot ^2 \frac{A}{2} \end{aligned} \end{aligned} $

Given that in a $\triangle A B C$

$a=13, b=14$ and $c=15$

we know that $r_1=\frac{\Delta}{s-a}$

where, $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

and $s=\frac{a+b+c}{2}$

So, $s=\frac{13+14+15}{2}=\frac{42}{2}=21$

$ \begin{aligned} Now,\Delta & =\sqrt{21 \times 8 \times 7 \times 6} \\ & =\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\ & =2 \times 2 \times 3 \times 7=84 \end{aligned} $

So, $\quad r_1=\frac{84}{21-13}=\frac{84}{8}=\frac{21}{2}$

Given that

$ r: R: r_2=1: 3: 7 $

we know that $A+B+C=180^{\circ}$

$ \begin{aligned} & Now,\,\,\,\,\sin (A+C)+\sin B \\ & =\sin \left(180^{\circ}-B\right)+\sin B \\ & =\sin B+\sin B=2 \sin B \end{aligned} $

$ \begin{aligned} & r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{\Delta}{s} \\ & R=\frac{a}{2 \sin A}=\frac{b}{2 \sin B}=\frac{C}{2 \sin C}=\frac{a b c}{4 \Delta} \end{aligned} $

On simplifying above ratio, we get

$ \begin{aligned} & B=90^{\circ} \\ & \therefore \sin (A+C)+\sin B=2 \sin 90^{\circ}=2 \times 1=? \end{aligned} $

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

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

Now, $\left(r_1+r_2\right) \operatorname{cosec}^2 \frac{C}{2}$

$ \begin{aligned} & =\left(\begin{array}{rl} 4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} +4 R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} \end{array}\right) \cdot \operatorname{cosec^2} \frac{C}{2}\\ & =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\cos \frac{A}{2} \sin \frac{B}{2}\right]{cosec^2} \frac{C}{2} \end{aligned} $

$ =4 R \cos \frac{C}{2}\left[\sin \left(\frac{A}{2}+\frac{B}{2}\right)\right] \operatorname{cosec}^2 \frac{C}{2} $

Given $\triangle A B C, A, B, C$ in $A P$ which means $B-C=A-B$ where $C>B>A$

$ \Rightarrow \quad 2 B=A+C $

Sum of angles of $\triangle A B C$

$ A+B+C=180^{\circ} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…(i)$

$\because A, B, C$ in AP, we can assume that $c=a$,

$ B=a+d, A=a+2 d $

From Eq. (i),

$ \begin{array}{r} a+2 d+a+d+a=180^{\circ} \\ 3 a+3 d=180 \\ a+d=\frac{180}{3} \Rightarrow B=60^{\circ} \end{array} $

Thus, $A+C=120^{\circ}$

Given $\cos A+\cos B+\cos C=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…(ii)$

Let $C=45^{\circ}$ (we need to check for angles)

Then, $B=60^{\circ}, A=75^{\circ}$

Using Eq. (ii),

$ \begin{aligned} & \cos 75^{\circ}+ \cos 60^{\circ}+\cos 45^{\circ}= \frac{\sqrt{3}-1}{2 \sqrt{2}}+\frac{1}{2}+\frac{1}{\sqrt{2}} \\ &= \frac{2+\sqrt{2}+\sqrt{3}-1}{2 \sqrt{2}}=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}} \end{aligned} $

This matches the expression.

Therefore, $A=75^{\circ}$ and $\tan A=\tan 75^{\circ}$

$ \tan A=2+\sqrt{3} $

In triangle $ \triangle ABC $, given the ratio of the sums of the sides as $ b+c : c+a : a+b = 7 : 8 : 9 $, we can express these sums using a common factor $ k $ as follows:

$ b+c = 7k $

$ c+a = 8k $

$ a+b = 9k $

By adding up these equations, we have:

$ b+c+c+a+a+b = 7k + 8k + 9k = 24k $

Simplifying, we get:

$ 2(a+b+c) = 24k \quad \Rightarrow \quad a+b+c = 12k $

Using this, we can solve for each side:

From $ a + 7k = 12k $, it follows that $ a = 5k $.

From $ c + a = 8k $, substitute $ a = 5k $ to find $ c = 3k $.

From $ a + b = 9k $, substitute $ a = 5k $ to find $ b = 4k $.

Thus, the side lengths are:

$ a = 5k $

$ b = 4k $

$ c = 3k $

Next, we apply the cosine rule to find the angle $ C $:

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

Substituting the values, we have:

$ \begin{aligned} \cos C & = \frac{(4k)^2 + (5k)^2 - (3k)^2}{2 \times 4k \times 5k} \\ & = \frac{16k^2 + 25k^2 - 9k^2}{40k^2} \\ & = \frac{32k^2}{40k^2} = \frac{4}{5} \end{aligned} $

Therefore, the smallest angle $ C $ in radians is:

$ C = \cos^{-1}\left(\frac{4}{5}\right) $

In the given triangle $ \triangle ABC $, we are provided with the equation:

$ (a+c)^2 = b^2 + 3ac $

Using the cosine rule for triangles:

$ \begin{aligned} a^2 &= b^2 + c^2 - 2bc \cos A \\ b^2 &= a^2 + c^2 - 2ac \cos B \\ c^2 &= a^2 + b^2 - 2ab \cos C \\ \end{aligned} $

Additionally, we know the sum of angles in a triangle is $ \pi $, i.e., $ A + B + C = \pi $.

Given the equation:

$ (a+c)^2 = b^2 + 3ac $

We rearrange this equation:

$ \begin{aligned} a^2 + c^2 + 2ac &= b^2 + 3ac \\ a^2 + c^2 - b^2 &= ac \quad \quad \quad \quad \,\,\,\,(i) \\ \end{aligned} $

From the cosine rule:

$ a^2 + c^2 - b^2 = 2ac \cos B \quad \quad \quad \,\,\,\,\,(ii) $

Equating Equation (i) and Equation (ii), we get:

$ \begin{aligned} ac &= 2ac \cos B \\ 1 &= 2 \cos B \\ \cos B &= \frac{1}{2} \end{aligned} $

Thus, $ \angle B = \frac{\pi}{3} $ or $ 60^\circ $.

Therefore, we have:

$ \frac{a+c}{2R} = \sin B \Rightarrow \sin 60^\circ = \frac{\sqrt{3}}{2} $

Thus:

$ \frac{a+c}{2R} = \sqrt{3} \cos \left(\frac{A-C}{2}\right) $

Given that the angles $\angle A, \angle B, \angle C$ in $\triangle ABC$ are in arithmetic progression, we can express $\angle B$ as the average of $\angle A$ and $\angle C$, i.e., $B = \frac{A + C}{2}$.

We also know that the sum of angles in a triangle is $180^\circ$:

$ A + B + C = 180^\circ $

Substituting $B = \frac{A + C}{2}$ into this equation gives:

$ A + \frac{A + C}{2} + C = 180^\circ $

$ 2A + A + C + 2C = 180^\circ \times 2 $

$ \frac{A + C}{2} = 60^\circ = B $

Therefore, $B = 60^\circ$.

The area of $\triangle ABC$ is given as $\frac{\sqrt{3}}{2}$.

Regarding the radii of the excircles, denoted as $r_1, r_2, r_3$ for the sides opposite to vertices $A, B, C$ respectively, the condition given is:

$ r_1 r_2 = r_2 r $

For any triangle, the area can also be expressed as:

$ \frac{abc}{4R} $

Thus, equating the given area:

$ \frac{\sqrt{3}}{2} = \frac{abc}{4R} $

Solving for $R$:

$ R = \frac{abc}{2\sqrt{3}} $

Since we have determined that $B = 60^\circ$, the circumradius $R$ in such a triangle is given by:

$ R = \frac{b}{\sqrt{3}} $

Calculating the circumradius directly for special triangles:

Hence, the circumradius $R$ is 1.

We have, two larger side of a triangle are 8 and 7 and smaller side is a and angles are in AP

According to the given condition.

$ \begin{aligned} \Rightarrow & & a+a-d+a+d & =180^{\circ} \\ \Rightarrow & & a & =60^{\circ} \end{aligned} $

$ \begin{aligned} & \text { Then, } \quad \cos 60^{\circ}=\frac{a^2+b^2-c^2}{2 a c} \\ & \Rightarrow \quad \frac{1}{2}=\frac{a^2+64-49}{2 \times a \times 8} \\ & \Rightarrow a^2-8 a+15=0 \end{aligned} $

On solving, we get

$ \begin{aligned} & \Rightarrow \quad a=\frac{8 \pm \sqrt{64-60}}{2}=\frac{8 \pm 2}{2} \\ & \Rightarrow \quad a_1=3 a_2=5 \end{aligned} $

Hence, the value satisfy the condition $a_1

Then, $2 a_1+3 a_2=2 \times 3+3 \times 5$

$ \Rightarrow \quad 2 a_1+3 a_2=21 $

Given the triangle $ \triangle ABC $ with sides $ a = 13 $, $ b = 14 $, and $ \cos \left(\frac{C}{2}\right) = \frac{3}{\sqrt{13}} $.

Step 1: Utilize the Cosine Rule

First, we find $\cos C$ using:

$ \cos C = 2\cos^2\left(\frac{C}{2}\right) - 1 $

Substitute the given value:

$ \cos C = 2 \left(\frac{3}{\sqrt{13}}\right)^2 - 1 = 2 \times \frac{9}{13} - 1 = \frac{18 - 13}{13} = \frac{5}{13} $

Now, apply the cosine rule to find $ c $:

$ c^2 = a^2 + b^2 - 2ab \cos C $

Substitute the known values:

$ c^2 = 13^2 + 14^2 - 2 \times 13 \times 14 \times \left(\frac{5}{13}\right) $

Calculate:

$ c^2 = 365 - 140 = 225 $

$ c = 15 $

Step 2: Calculate the Semi-perimeter

The semi-perimeter $ S $ is:

$ S = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 $

Step 3: Find the Area using Heron's Formula

Calculate the area $\Delta$:

$ \Delta = \sqrt{S(S-a)(S-b)(S-c)} = \sqrt{21 \times (21-13) \times (21-14) \times (21-15)} $

$ \Delta = \sqrt{21 \times 8 \times 7 \times 6} = 84 $

Step 4: Calculate $ r_1 $

The radius $ r_1$ of the excircle opposite $ A $ is:

$ r_1 = \frac{\Delta}{S-a} = \frac{84}{21-13} = \frac{84}{8} = 10.5 $

Multiplying by 2:

$ 2r_1 = 2 \times 10.5 = 21 $

Conclusion

From the calculations, we find that $ 2r_1 = S $.

$ \begin{aligned} &\text { Given, in a } \triangle A B C\\ &\begin{array}{rrr} & \quad\left(r_2-r_1\right)\left(r_3-r_1\right)=2 r_2 r_3 \\ \Rightarrow & \left(r_1 r_2+r_2 r_3+r_1 r_3\right)-r_1^2=0 \\ \Rightarrow & \frac{r_1 r_2 r_3}{r}-r_1^2=0 \\ \Rightarrow & r_1\left(\frac{r_2 r_3}{r}-r_1\right)=0 \\ \Rightarrow & \quad m_1=r_2 r_3 \quad\left(\because r_1 \neq 0\right) \end{array} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{\Delta}{s} \cdot \frac{\Delta}{s-a}=\frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} \\ \Rightarrow & \frac{(s-b)(s-c)}{s(s-a)}=1 \\ \Rightarrow & \tan ^2 \frac{A}{2}=\tan ^2 45^{\circ} \\ \Rightarrow & A=90^{\circ} \\ \therefore & \frac{B+C}{2}=45^{\circ} \end{array}\\ &\text { Now, } 2(r+R)=2\left(4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}+R\right)\\ &\begin{aligned} &=2 R\left(4\left(\frac{1}{\sqrt{2}}\right) \sin \frac{B}{2} \sin \frac{C}{2}+1\right) \\ &=2 \sqrt{2} R\left(2 \sin \frac{B}{2} \sin \frac{C}{2}+\frac{1}{\sqrt{2}}\right) \\ &=2 \sqrt{2} R\left(2 \sin \frac{B}{2} \sin \frac{C}{2}+\cos \left(\frac{B+C}{2}\right)\right) \end{aligned} \end{aligned} $

$ \begin{aligned} & =2 \sqrt{2} R\left(2 \sin \frac{B}{2} \sin \frac{C}{2}+\cos \frac{B}{2} \cos \frac{C}{2}-\sin \frac{B}{2} \sin \frac{C}{2}\right) \\ & =2 \sqrt{2} R\left(\cos \frac{B}{2} \cos \frac{C}{2}+\sin \frac{B}{2} \sin \frac{C}{2}\right)=2 \sqrt{2} R \cos \left(\frac{B-C}{2}\right) \end{aligned} $

Given, angle ratio is $3: 2: 1$ We need to find sides Ratio.

$ \begin{aligned} & \text { } \begin{aligned} Let \,A=3 \theta, B & =2 \theta \text { and } C=\theta \\ \because \quad A+B+C & =\pi \quad \text { [angle sum property] } \\ 3 \theta+2 \theta+\theta & =\pi \\ 6 \theta & =\pi \\ \theta & =\frac{\pi}{6} \\ A & =3 \times \frac{\pi}{6}=\frac{\pi}{2} \\ B & =2 \times \frac{\pi}{6}=\frac{\pi}{3} \\ C & =\frac{\pi}{6} \\ a: b: c & =\sin \frac{\pi}{2}: \sin \frac{\pi}{3}: \sin \frac{\pi}{6} \\ \Rightarrow \quad a: b: c & =1: \frac{\sqrt{3}}{2}: \frac{1}{2} \\ a: b: c & =2: \sqrt{3}: 1 \end{aligned} \end{aligned} $

In $\triangle A B C, B C=5, C A=6, A B=7$.

Draw median from $B$ onto $A C=1$.

In $\triangle A B C$, point $D$ is mid-point of $A C$.

$ \begin{aligned} A D & =D C=\frac{1}{2} A C=3 \\ (A B)^2+(B C)^2 & =2 B D^2+2 D A^2 \end{aligned} $

[using Apollonius theorem]

$ \begin{aligned} (7)^2+(5)^2 & =2 B D^2+2 \times(3)^2 \\ 49+25 & =2 B D^2+2 \times 9 \\ 74 & =2 B D^2+18 \\ 2 B D^2 & =74-18=56 \\ B D^2 & =28 \\ B D^2 & =7 \times 4 \Rightarrow B D=2 \sqrt{7} \end{aligned} $

Hence, length of median drawn from $B$ onto

$ A C=2 \sqrt{7} $

To find the ratio $ R:r $ in $\triangle ABC$ with side ratios $ AB:BC:CA = 6:4:5 $, we use the following formula:

$ \frac{R}{r} = \frac{abc}{4(s-a)(s-b)(s-c)} $

where $ s $ is the semi-perimeter, which can be calculated as:

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

Given the side ratios, let $ a = 6x $, $ b = 4x $, and $ c = 5x $. Therefore, the semi-perimeter $ s $ is:

$ s = \frac{6x + 4x + 5x}{2} = \frac{15x}{2} $

Substitute into the formula:

$ \frac{R}{r} = \frac{6x \cdot 4x \cdot 5x}{4 \left( \frac{15x}{2} - 6x \right) \left( \frac{15x}{2} - 4x \right) \left( \frac{15x}{2} - 5x \right)} $

Calculate each term:

$ s-a = \frac{15x}{2} - 6x = \frac{15x - 12x}{2} = \frac{3x}{2} $

$ s-b = \frac{15x}{2} - 4x = \frac{15x - 8x}{2} = \frac{7x}{2} $

$ s-c = \frac{15x}{2} - 5x = \frac{15x - 10x}{2} = \frac{5x}{2} $

Substitute back in:

$ \frac{R}{r} = \frac{6x \cdot 4x \cdot 5x}{4 \cdot \left(\frac{3x}{2}\right) \cdot \left(\frac{7x}{2}\right) \cdot \left(\frac{5x}{2}\right)} $

Simplify:

$ \frac{R}{r} = \frac{120x^3}{4 \cdot \frac{3x}{2} \cdot \frac{7x}{2} \cdot \frac{5x}{2}} = \frac{120x^3}{4 \cdot \frac{105x^3}{8}} $

$ = \frac{120x^3 \cdot 8}{4 \cdot 105x^3} = \frac{960}{420} = \frac{16}{7} $

Thus, $ R:r $ is $ \frac{16}{7} $.

Altitude of $B C$,

$ \begin{aligned} y-2 & =\frac{1}{3}(x-2) \\ 3 y-6 & =x-2 \\ x-3 y+4 & =0 \end{aligned} $

Altitude of $A C, y+1=-1(x-5)$

$ \begin{aligned} y-1 & =-x+5 \\ x+y-6 & =0 \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} -4 y+10 & =0 \\ y & =+\frac{5}{2} \\ \Rightarrow \quad \beta & =\frac{5}{2} \Rightarrow x+\frac{5}{2}=6 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2 x=12-5 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{7}{2} \\ & \,\,\,\,\,\,\,\,\,\,\,\alpha=\frac{7}{2} \\ & \Rightarrow \quad \alpha+\beta=\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\ & \therefore \quad \alpha+\beta=6 \end{aligned} $

$\because \sin ^2 \frac{A}{2}=\frac{(s-b)(s-c)}{b c}$,

$ \sin ^2 \frac{B}{2}=\frac{(s-c)(s-a)}{a c} $

and $\sin ^2 \frac{C}{2}=\frac{(s-a)(s-b)}{a b}$

$\Rightarrow \sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}$

$=\frac{(s-b)(s-c)}{b c}+\frac{(s-c)(s-a)}{a c}+\frac{(s-a)(s-b)}{a b} \quad \ldots \text { (i) } $

Given, $4 r_1=5 r_2=6 r_3$

$\Rightarrow r_1: r_2=5: 4$ and $r_2: r_3=6: 5$

$\Rightarrow r_1: r_2: r_3=30: 24: 20$

$\Rightarrow \frac{s}{s-a}: \frac{s}{s-b}: \frac{s}{s-c}=30: 24: 20$

$\Rightarrow s-a: s-b: s-c=\frac{1}{30}: \frac{1}{24}: \frac{1}{20}$

$\therefore s-b+s-c=2 s-b-c=a$

$ =\frac{1}{24}+\frac{1}{20}=\frac{11}{120} $

Similarly, $b=\frac{1}{30}+\frac{1}{20}=\frac{50}{600}=\frac{10}{120}$

and $\quad C=\frac{1}{30}+\frac{1}{24}=\frac{54}{720}=\frac{9}{120}$

$\therefore \quad a: b: c=11: 10: 9$

$s=\frac{a+b+c}{2}=\frac{11+10+9}{2}=15$

From Eq. (i), we get

$ \begin{aligned} \frac{\sin ^2 A}{2} & +\frac{\sin ^2 B}{2}+\frac{\sin ^2 C}{2} \\ & =\frac{5 \times 6}{90}+\frac{6 \times 4}{99}+\frac{4 \times 5}{110} \\ \Rightarrow \quad & \frac{30}{90}+\frac{24}{99}+\frac{20}{110}=\frac{25}{33} \end{aligned} $

We know that

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

$\cot \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$,

$\cot \frac{B}{2}=\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$

$\cot \frac{C}{2}=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}, r=\frac{\Delta}{s}$

Now, $\pi_1 \cot \frac{A}{2}+\pi_2 \cot \frac{B}{2}+\pi_3 \cot \frac{C}{2}$

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

$ \begin{aligned} & =\Delta^2 \sqrt{\frac{1}{s(s-a)(s-b)(s-c)}} +\Delta^2 \sqrt{\frac{1}{(s-a)(s-b)(s-c) s}} +\Delta^2 \sqrt{\frac{1}{(s-a)(s-b)(s-c)_1}} \\ & =3 \Delta^2 \cdot \frac{1}{\sqrt{s(s-a)(s-b)(s-c)}}=3\Delta \\ & rr_1 \cot \frac{A}{2}+rr_2 \cot \frac{B}{2}+rr_3 \cot \frac{C}{2}=3 \Delta \end{aligned} $

$ \begin{aligned} \triangle A B C, b c-r_2 r_3 & =b c-\frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} \\ & =b c-\frac{\Delta^2}{(s-b)(s-c)} \\ & =b c-\frac{s(s-a)(s-b)(s-c)}{(s-b)(s-c)} \\ & =b c-s(s-a)=(b c-s(s-a)) \\ & =b c\left[1-\frac{s(s-a)}{b c}\right] \\ & =b c\left(1-\cos ^2 \frac{A}{2}\right)=b c \cdot \sin ^2 \frac{A}{2} \\ & =b c \cdot \frac{(s-b)(s-c)}{b c} \\ & =\frac{(s-b)(s-c)(s-a) \cdot s}{s(s-a)} \\ & =\frac{\Delta^2}{s(s-a)}=\frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \\ & =r \cdot r_1 \end{aligned} $