Properties of Triangles

$ \begin{gathered} A D=\frac{2 b c}{b+c} \cos \frac{A}{2} \\ \Rightarrow D E \cos \frac{A}{2}=\frac{D E \times A D \times(b+c)}{2 b c} \\ =\frac{B D \times D C \times(b+c)}{2 b c} \\ =\frac{a c}{b+c} \times \frac{a b}{b+c} \times \frac{(b+c)}{2 b c} \\ =\frac{a^2}{2(b+c)} \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \tan \left(\frac{C}{2}\right)=\frac{\Delta}{s(s-c)} \\ & r=(s-c) \tan \left(\frac{c}{2}\right) \\ & \text { Also, } s-c=\frac{a b \cos ^2\left(\frac{c}{2}\right)}{s} \\ & \Rightarrow \quad r=\frac{a b}{s} \cos ^2\left(\frac{c}{2}\right) \tan \left(\frac{c}{2}\right) \\ & \Rightarrow \quad r=\frac{5 \times 4}{s \times\left(1+\frac{7}{9}\right)} \times \frac{\sqrt{7}}{3} \\ & \Rightarrow \quad r=\frac{5 \times 4 \times 9 \times \sqrt{7}}{3 \times 16 \times s} \\ & \Rightarrow \quad r=\frac{15 \sqrt{7}}{4 s} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { As } \tan \left(\frac{C}{2}\right)=\sqrt{\frac{7}{9}} \Rightarrow \cos \left(\frac{C}{2}\right)=\frac{3}{4} \\ & \cos C=2 \cos ^2\left(\frac{C}{2}\right)-1=\frac{2 \times 9}{16}-1=\frac{1}{8} \\ & \Rightarrow \quad \cos C=\frac{a^2+b^2-c^2}{2 a b} \\ & \Rightarrow \quad \frac{1}{8}=\frac{41-c^2}{40} \\ & \Rightarrow \quad c^2=36 \Rightarrow c=6 \\ & \Rightarrow \quad s=\frac{a+b+c}{2}=\frac{9+6}{2} \\ & \Rightarrow r=\frac{15 \sqrt{7}}{4 \times 15} \times 2=\frac{\sqrt{7}}{2} \end{aligned} $

$ \text { Let equation of } B C \text { be } x-y=0 \text {. } $

Reflection of $H(5,8)$ about side $B C$ will be point $P(8,5)$

$ \begin{aligned} \text { Radius } & =O P \\ & =\sqrt{(8-2)^2+(5-3)^2} \\ & =\sqrt{36+4}=2 \sqrt{10} \end{aligned} $

$ \begin{aligned} & =\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C} \\ & =\frac{2 R \cos A}{a}+\frac{2 R \cos B}{b}+\frac{2 R \cos C}{c} \\ & =2 R\left(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}\right) \\ & =2 R\left(\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2 a b c}\right) \\ & =\frac{R}{a b c}\left(a^2+b^2+c^2\right) \\ & =\frac{a^2+b^2+c^2}{4 \Delta} \quad(\Delta=\text { area of triangle }) \end{aligned} $

Now, $s=\frac{a+b+c}{2}=\frac{15}{2}$

$ \begin{aligned} s-a & =\frac{9}{2}, s-b=\frac{5}{2} \text { and } s-c=\frac{1}{2} \\ \Delta & =\sqrt{s \times(s-a) \times(s-b) \times(s-c)} \\ & =\sqrt{\frac{15}{2} \times \frac{9}{2} \times \frac{5}{2} \times \frac{1}{2}} \\ & =\frac{15 \sqrt{3}}{4} \end{aligned} $

$ \begin{aligned} \Rightarrow \cot A+\cot B+\cot C & =\frac{3^2+5^2+7^2}{4 \times \frac{15}{4} \times \sqrt{3}} \\ & =\frac{83}{15 \sqrt{3}} \end{aligned} $

$ \begin{gathered} p_1=\frac{2 \Delta}{a}, \quad p_2=\frac{2 \Delta}{b}, \quad p_3=\frac{2 \Delta}{c} \\ \frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=\frac{a^2+b^2+c^2}{4 \Delta^2} \\ r_1=\frac{\Delta}{s-a^{\prime}}, \quad r_2=\frac{\Delta}{s-b^{\prime}}, \quad r_3=\frac{\Delta}{s-c} \\ \Rightarrow s-a=\frac{\Delta}{4}, \quad s-b=\frac{\Delta}{6}, \quad s-c=\frac{\Delta}{12} \\ \Rightarrow b+c-a=\frac{\Delta}{2}, \quad a+c-b=\frac{\Delta}{3}, \\ a+b-c=\frac{\Delta}{6} \\ \Rightarrow a+b+c=\Delta\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=\Delta(1) \\ \Rightarrow a=\frac{\Delta}{4}, \quad b=\frac{\Delta}{3}, \quad c=\frac{5 \Delta}{12} \\ \Rightarrow \frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=\frac{1}{4}\left(\frac{1}{16}+\frac{1}{9}+\frac{25}{144}\right) \\ =\frac{1}{4} \times \frac{9+16+25}{144} \\ =\frac{50}{4 \times 144}=\frac{25}{288} \end{gathered} $

$ \begin{aligned} &\begin{aligned} & a=3, b=5, c=7, s=\frac{15}{2} \\ & \Delta=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{\frac{15}{2}\left(\frac{15}{2}-3\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-7\right)} \\ &=\frac{1}{4} \sqrt{15 \times 9 \times 5 \times 1}=\frac{5 \times 3}{4} \sqrt{3} \\ & \Rightarrow \quad \frac{15}{4} \sqrt{3} \end{aligned}\\ &\begin{aligned} \operatorname{Circum} \operatorname{radius}(R) & =\frac{a b c}{4 \Delta} \\ & =\frac{3 \times 5 \times 7}{4\left(\frac{15}{4} \sqrt{3}\right)}=\frac{7}{\sqrt{3}} \end{aligned} \end{aligned} $

Step 1: Find how far each ship goes in 2 hours.

The first ship travels at 8 km per hour. In 2 hours, it travels:
$8 \times 2 = 16$ km

The second ship travels at 12 km per hour. In 2 hours, it travels:
$12 \times 2 = 24$ km

Step 2: Find the angle between the directions of the ships.

The paths of the two ships make a $120^\circ$ angle with each other. We use this angle to find the distance between them after 2 hours.

Step 3: Use the Cosine Law to find the distance between ships.

The Cosine Law formula is: $c^2 = a^2 + b^2 - 2ab\cos{C}$ where $a$ and $b$ are the distances traveled, and $C$ is the angle between the paths.

Let $a = 24$ km, $b = 16$ km, and $C = 120^\circ$.

Plug these values into the formula:

$ \cos 120^\circ = \frac{24^2 + 16^2 - c^2}{2 \times 24 \times 16} $

Since $\cos 120^\circ = -\frac{1}{2}$, $ -\frac{1}{2} \times 2 \times 24 \times 16 = 24^2 + 16^2 - c^2 $

$24^2 = 576$, $16^2 = 256$, and $2 \times 24 \times 16 = 768$. So, $ -\frac{1}{2} \times 768 = 576 + 256 - c^2 \\ -384 = 832 - c^2 $

Add $c^2$ on both sides and add 384 to both sides: $ c^2 = 832 + 384 = 1216 $

Now, $1216 = 64 \times 19$. So, $ c = 8 \sqrt{19} $

So, the distance between the ships after 2 hours is $8\sqrt{19}$ km.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } c^2-a^2=\sqrt{3} b c-b^2 \\ & \frac{b^2+c^2-a^2}{2 b c}=\frac{\sqrt{3} b c}{2 b c} \\ & \cos A=\frac{\sqrt{3}}{2} \\ & \Rightarrow \quad A=30^{\circ} \\ & b^2-a^2=c^2-a c \\ & c^2+a^2-b^2=a c \end{aligned} \end{aligned} $

$ \begin{array}{ll} & \frac{c^2+a^2-b^2}{2 a c}=\frac{1}{2} \\ & \cos B=\frac{1}{2} \\ \Rightarrow \quad & B=60^{\circ} \\ \because \quad & \angle A+\angle B+\angle C=180^{\circ} \\ & \angle C=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ} \end{array} $

$ \text { Given, } \angle B=90^{\circ}, a=13, c=84 $

$ \begin{aligned} b^2 & =a^2+c^2 \\ b^2 & =7225 \\ b & =85 \\ \therefore \quad r & =\frac{\Delta}{s} \\ & =\frac{\frac{1}{2} \times 13 \times 84}{\frac{13+84+85}{2}} \\ & =\frac{13 \times 84}{182} \\ & =\frac{13 \times 6}{13}=6 \\ R & =\frac{b}{2}=\frac{85}{2}=42.5 \\ \therefore r+R & =6+425=48.5 \end{aligned} $

$ \begin{aligned} &\text { We have, } r_1=4, r_2=8 \text { and } r_3=24\\ &\therefore \quad 4=\frac{\Delta}{s-a}, 8=\frac{\Delta}{s-b} \text { and } 24=\frac{\Delta}{s-c} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad s-a=\frac{\Delta}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \qquad s-b=\frac{\Delta}{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ & \text { And } \quad s-c=\frac{\Delta}{24} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

On adding these three equations, we get

$ \begin{aligned} & 3 s-(a+b+c)=\frac{\Delta}{4}+\frac{\Delta}{8}+\frac{\Delta}{24} \\ \Rightarrow & 3 s-2 s=\frac{6 \Delta+3 \Delta+\Delta}{24}=\frac{10 \Delta}{24} \Rightarrow s=\frac{5}{12} \Delta \end{aligned} $

From Eq. (i), $a=s-\frac{\Delta}{4}=\frac{5 \Delta}{12}-\frac{\Delta}{4}=\frac{2 \Delta}{12}=\frac{\Delta}{6}$

From Eq. (ii), $b=s-\frac{\Delta}{8}=\frac{5}{12} \Delta-\frac{\Delta}{8}=\frac{7}{24} \Delta$

From Eq. (iii), $c=s-\frac{\Delta}{24}=\frac{5}{12} \Delta-\frac{\Delta}{24}=\frac{9}{24} \Delta$

$ \therefore a: b: c=\frac{\Delta}{6}: \frac{7}{24} \Delta: \frac{9}{24} \Delta=4: 7: 9 $

$ \begin{aligned} &\text { We know that }\\ &\begin{array}{ll} & 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} \\ \therefore \quad & \left(r_2+r_3\right)=4 R \cos \frac{A}{2}\left[\sin \frac{B}{2} \cos \frac{C}{2}+\cos \frac{B}{2} \sin \frac{C}{2}\right] \\ \Rightarrow \quad & \left(r_2+r_3\right)=4 R \cos \frac{A}{2} \sin \left(\frac{B+C}{2}\right) \\ \Rightarrow \quad & \left(r_2+r_3\right)=4 R \cos \frac{A}{2} \sin \left(90^{\circ}-\frac{A}{2}\right)=4 R \cos \frac{A}{2} \cos \frac{A}{2} \end{array} \end{aligned} $

$ \begin{aligned} \therefore\left(r_2+r_3\right) \operatorname{cosec}^2\left(\frac{A}{2}\right) & =4 R \cos ^2 \frac{A}{2} \operatorname{cosec}^2 \frac{A}{2} \\ & =\frac{4 R}{\tan ^2 \frac{A}{2}} \end{aligned} $

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

Since, $p_1, p_2$ and $p_3$ are altitudes of the sides $a, b$ and $c$ respectively.

$ \begin{array}{ll} \therefore & p=\frac{2 \Delta}{\text { base }} \\ \Rightarrow & p_1=\frac{2 \Delta}{a}, p_2=\frac{2 \Delta}{b}, p_3=\frac{2 \Delta}{c} \\ \Rightarrow & \frac{1}{p_1^2}=\frac{a^2}{4 \Delta^2}, \frac{1}{p_2^2}=\frac{b^2}{4 \Delta^2}, \frac{1}{p_3^2}=\frac{c^2}{4 \Delta^2} \\ \therefore & \frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2} \\ & =\frac{a^2}{4 \Delta^2}+\frac{b^2}{4 \Delta^2}+\frac{c^2}{4 \Delta^2} \\ & =\frac{a^2+b^2+c^2}{4 \Delta^2} \end{array} $

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

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

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

(Using heron's formula)

$ \begin{aligned} & =\sqrt{\frac{15}{2} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2}} \\ & =\sqrt{\frac{1575}{16}}=\frac{\sqrt{1575}}{4} \\ & =\Delta^2=\frac{1575}{16} \end{aligned} $

And $a^2+b^2+c^2=4^2+5^2+6^2$

$ =16+25+36=77 $

$ \begin{aligned} So,\,\,& \frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=\frac{77}{4 \Delta^2} \\ & =\frac{77}{4 \times \frac{1575}{16}}=\frac{77}{\frac{6300}{16}} \\ & =\frac{77 \times 16}{6300}=\frac{44}{225} \end{aligned} $

Given angles $A, B, C$ are in AP and $A+B+C=180^{\circ} \quad(\because$ angles of $\triangle A B C)$

So, $\frac{A+C}{2}=B$

$ \therefore 2 B+B=180^{\circ} $

$ \begin{array}{lrl} \Rightarrow & & 3 B=180^{\circ} \\ \Rightarrow & & B=60^{\circ} \end{array} $

Given that $r_3^2=r_1 r_2+r_2 r_3+r_3 r_1$

Since, $B=60^{\circ}$, using the cosine rule for side $b$.

$ \begin{aligned} b^2 & =a^2+c^2-2 a c \cos 60^{\circ} \\ & =a^2+c^2-2 a c \cdot \frac{1}{2}=a^2+c^2-a c \end{aligned} $

We know that, $s^2=r_1 r_2+r_2 r_3+r_3 r_1$

And given, $r_3^2=r_1 r_2+r_2 r_3+r_3 r_1$

So, $r_3^2=s^2$

$ \Rightarrow r_3=s $

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

$ \begin{aligned} & \Rightarrow \quad s=\frac{\Delta}{s-c} \Rightarrow \Delta=s(s-c) \\ & \Rightarrow \quad \sqrt{s(s-a)(s-b)(s-c)}=s(s-c) \end{aligned} $

Squaring both sides, we get

$ \begin{array}{ll} \Rightarrow & s(s-a)(s-b)(s-c)=s^2(s-c)^2 \\ \Rightarrow & (s-a)(s-b)=s(s-c) \\ \Rightarrow & s^2-s a-s b+a b=s^2-s c \\ \Rightarrow & a b=s(a+b-c) \\ \Rightarrow & a b=\frac{(a+b+c)}{2} \cdot(a+b-c) \\ & {\left[\because s=\frac{a+b+c}{2}\right]} \end{array} $

$ \begin{array}{ll} \Rightarrow & 2 a b=(a+b)^2-c^2 \\ \Rightarrow & 2 a b=a^2+b^2+2 a b-c^2 \\ \Rightarrow & a^2+b^2-c^2=0 \\ \Rightarrow & c^2=a^2+b^2 \end{array} $

So, the triangle is a right-angled triangle with right angle at $c$.

But, we have $B=60^{\circ}$ and $C=90^{\circ}$

So, $A=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}$

Using the sine rule,

$ \begin{aligned} & & \frac{a}{\sin A} & =\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow & & \frac{a}{\sin 30^{\circ}} & =\frac{b}{\sin 60^{\circ}} \\ \Rightarrow & & \frac{a}{\frac{1}{2}} & =\frac{b}{\frac{\sqrt{3}}{2}} \Rightarrow 2 a=\frac{2 b}{\sqrt{3}} \\ \Rightarrow & & \sqrt{3} a & =b \\ \therefore & & b & =\sqrt{3} a \end{aligned} $

$b=3, c=\sqrt{19}$ 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} $

$ a=5, b=3, c=7 $

$ \cos A=\frac{9+49-25}{2 \cdot 3 \cdot 7}=\frac{11}{14} $

$ \cos B=\frac{25+49-9}{2 \cdot 5 \cdot 7}=\frac{65}{2 \cdot 5 \cdot 7}=\frac{13}{14} $

$ \frac{\sin (A-B)}{\sin (A+B)}=\frac{\sin A \cos B-\cos A \cdot \sin B}{\sin (\pi-C)} $

$ =\frac{\sin A \cos B-\cos A \sin B}{\sin C} $

$ =\frac{a \cos B-b \cos A}{c}=\frac{5 \times \frac{13}{14}-3 \times \frac{11}{14}}{7} $

$ =\frac{65-33}{14 \times 7}=\frac{32}{14 \times 7}=\frac{16}{49} $

$ \Rightarrow \sqrt{\frac{\sin (A-B)}{\sin (A+B)}}=\sqrt{\frac{16}{49}}=\frac{4}{7} $

$ \begin{array}{l} r_{1}=6, r_{2}=9, r_{3}=18 \text {, } \\\\ \frac{r_{1}}{r_{2}}=\frac{s-b}{s-a}=\frac{6}{9}=\frac{2}{3} \\\\ \frac{r_{2}}{r_{3}}=\frac{s-c}{s-b}=\frac{1}{2}, \frac{r_{3}}{r_{1}}=\frac{s-a}{s-c}=3 \\\\ \Rightarrow \quad \frac{a+c-b}{b+c-a}=\frac{2}{3}, \frac{s-c}{s-b}=\frac{1}{2} \text {, } \frac{s-a}{s-c} =3 \\\\ \Rightarrow \quad \frac{a+c-b}{b+c-a} =\frac{2}{3}, \frac{a+b+c}{a+c-b}=\frac{1}{3} \frac{b+c-a}{a+b-c} =3 \\\\ \Rightarrow \quad 4 a+2 b =4 c \\\\ \Rightarrow \quad 2 a+b =2 c \quad \ldots \text { (i) } \\\\ \Rightarrow \quad 5 a+c =5 b \quad \ldots \text { (ii) } \\\\ a+3 b =3 c \quad \ldots \text { (iii) } \end{array} $

On solving, we get

$ a: b: c=3: 4: 5 $

$ \angle C=\frac{\pi}{2} \Rightarrow \cos A=\frac{4}{5} $

Given that $ \triangle ABC $ is an isosceles triangle with base $ BC $, we want to find $ r_1 $.

The known formulas are:

$ r = \frac{\Delta}{S} $

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

Where $ r $ is the inradius, $ r_1 $ is the exradius opposite the side $ a $, $\Delta$ is the area of the triangle, and $ S $ is the semi-perimeter.

Let's explore the product $ r \cdot r_1 $:

$ r \cdot r_1 = \frac{\Delta^2}{s(s-a)} = \frac{s(s-a)(s-b)(s-c)}{s(s-a)} $

Since the triangle is isosceles with $ b = c $:

$ r \cdot r_1 = (s-b)(s-c) = (s-b)^2 $

In the context of an isosceles triangle:

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

Since $ b = c $, we have:

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

Thus:

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

Rewriting this in terms of the circumcircle radius $ R $:

$ \frac{a}{\sin A} = 2R $

Therefore:

$ r \cdot r_1 = \frac{a^2}{4} = \frac{4R^2 \sin^2 A}{4} = R^2 \sin^2 A $

Hence, for the isosceles triangle $ \triangle ABC $ with base $ BC $, we find:

$ r_1 = R^2 \sin^2 A $

Given data indicates:

$ r_1 + r_2 = 3R $

$ r_2 + r_3 = 2R $

Let's analyze these conditions:

From $ r_1 + r_2 = 3R $:

We start by substituting in the formula:

$ \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{3abc}{4\Delta} $

Simplifying, we have:

$ \Delta^2 \left( \frac{1}{s-a} + \frac{1}{s-b} \right) = \frac{3abc}{4} $

Using the identity:

$ s(s-b)(s-c) + s(s-a)(s-c) = \frac{3abc}{4} $

Combining terms:

$ s(s-c)(s-b+s-a) = \frac{3abc}{4} $

Further simplification:

$ s(s-c)c = \frac{3abc}{4} $

Resulting relationship:

$ s(s-c) = \frac{3ab}{4} $

Therefore:

$ \sqrt{\frac{s(s-c)}{ab}} = \frac{\sqrt{3}}{2} $

Which implies:

$ \cos\left(\frac{C}{2}\right) = \frac{\sqrt{3}}{2} $

Hence,

$ \frac{C}{2} = 30^\circ \quad \text{or} \quad C = 60^\circ $

From $ r_2 + r_3 = 2R $:

Use:

$ s(s-a) = \frac{2bc}{4} $

Gives:

$ \sqrt{\frac{s(s-a)}{bc}} = \frac{1}{\sqrt{2}} $

Therefore:

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

Thus,

$ \frac{A}{2} = 45^\circ \quad \text{or} \quad A = 90^\circ $

Final Angles of the Triangle:

$\angle A = 90^\circ$

$\angle B = 60^\circ$

$\angle C = 30^\circ$

This implies the triangle does not fit typical ratios like $a:b:c = 2:1:\sqrt{3}$; instead, the lengths of sides must be inconsistent, i.e., $a \neq b \neq c$, which corresponds to a non-equilateral, non-isosceles right triangle.

In a triangle $ \triangle ABC $, consider the scenario where the sides $ b $ and $ c $ are fixed. When measuring angle $ A $, if there is a small error denoted by $ \delta A $, we wish to find the percentage error in the length of side $ a $.

The cosine rule for angle $ A $ is given by:

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

Rearranging, we have:

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

Taking the derivative with respect to angle $ A $ introduces an error term:

$ -2bc \sin A \cdot \delta A = -2a \cdot \delta a $

Simplifying gives:

$ bc \sin A \cdot \delta A = a \cdot \delta a $

Thus, the infinitesimal change in $ a $ is:

$ \delta a = \frac{bc \sin A \cdot \delta A}{a} $

Using the sine rule:

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

This implies:

$ a = 2R \sin A $

To find the percentage error in $ a $, we calculate:

$ \frac{\delta a}{a} \times 100 = \frac{\frac{bc \sin A \cdot \delta A}{a}}{a} \times 100 $

Substituting for $ a $:

$ = \frac{\frac{1}{2}bc \sin A \cdot \delta A}{\frac{1}{2} (2R \sin A)^2} \times 100 $

This simplifies to:

$ = \frac{\Delta \cdot \delta A}{2R^2 \sin^2 A} \times 100 $

where $ \Delta = \frac{1}{2} bc \sin A $.

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

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

$ =\frac{16+9-4}{2 \times 4 \times 3}=\frac{21}{24}=\frac{7}{8} $

$ \begin{array}{l} \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 \times a \times c}=\frac{16+4-9}{2 \times 4 \times 2}=\frac{11}{16} \\\\ \begin{array}{l} 2(a-b \cos C)(a-(c \sec B)) \\\\ 2\left(4-\frac{3 \times 4}{8}\right)\left(4-\frac{2 \times 16}{11}\right) \\\\ =2\left(4-\frac{21}{8}\right)\left(4-\frac{32}{11}\right) \\\\ =2\left(\frac{32-21}{8}\right)\left(\frac{44-32}{11}\right)=\frac{11}{4} \times \frac{12}{11}=3 \end{array} \end{array} $

Given, If $ A=45^{\circ}, C=75^{\circ} $,

$ R=\sqrt{2} $, then $ r= $ ?

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

$ \frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 75^{\circ}}=2 \sqrt{2} $

$ a \sqrt{2}=\frac{2 b}{\sqrt{3}}=\frac{2 \sqrt{2} c}{\sqrt{3}+1}=2 \sqrt{2} $

$ a=2, b=\sqrt{6}, c=\sqrt{3}+1 $

$ a+b+c=2 s \Rightarrow s=\frac{3+\sqrt{3}+\sqrt{6}}{2} $

$ \therefore r=(s-b) \tan \frac{B}{2} $

$ =\left[\frac{3+\sqrt{3}+\sqrt{6}}{2}-\sqrt{6}\right] \tan 30^{\circ} $

$ =\frac{3+\frac{\sqrt{3}-\sqrt{6}}{2 \sqrt{3}}=\frac{\sqrt{3}[\sqrt{3}+1-\sqrt{2}]}{2 \sqrt{3}}}{2} $

$ =\frac{(\sqrt{3}+1-\sqrt{2})(\sqrt{3}+1+\sqrt{2})}{2[\sqrt{3}+1+\sqrt{2}]} $

$ =\frac{3+1+2 \sqrt{3}-2}{2(\sqrt{3}+1+\sqrt{2})}=\frac{(\sqrt{3}+1)}{(\sqrt{3}+\sqrt{2}+1)} $