Properties of Triangles

Given, $ A \equiv(1,2,-3), B \equiv(2,3,-1) $

$ \begin{array}{l} C \equiv(3,1,1) \\\\ A B=\sqrt{1+1+4}=\sqrt{6}=c \\\\ A C=\sqrt{4+1+16}=\sqrt{21}=b \\\\ B C=\sqrt{1+4+4}=\sqrt{9}=a \\\\ \left|\frac{\cos A}{\cos B}\right|=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{\frac{a^{2}+c^{2}-b^{2}}{2 a c}}=\left|\frac{a\left(b^{2}+c^{2}-a^{2}\right)}{b\left(a^{2}+c^{2}-b^{2}\right)}\right| \\\\ =\left|\frac{\sqrt{9}}{\sqrt{21}} \frac{(21+6-9)}{(9+6-21)}\right|=\left|\frac{\sqrt{3}}{\sqrt{7}} \times \frac{18}{(-6)}\right| \end{array} $

Therefore, $ \left|\frac{\cos A}{\cos B}\right|=\frac{3 \sqrt{3}}{\sqrt{7}} $

Given, $A+B+C=2 S$

$ \begin{aligned} & \text { Now, } \sin (S-A) \cos (S-B)+\sin (S-C) \cos S \\\\ & =\frac{1}{2}[2 \sin (S-A) \cos (S-B)+2 \sin (S-C) \cos S] \\\\ & =\frac{1}{2}[\sin (2 S-A-B)+\sin (B-A) \\\\ & \quad+\sin (2 S-C)+\sin (-C)] . \\\\ & =\frac{1}{2}[\sin C-\sin (A-B)+\sin (A+B)-\sin C] \\\\ & =\frac{1}{2}[\sin (A+B)-\sin (A-B)] \\\\ & =\frac{1}{2}(2 \cos A \sin B)=\cos A \sin B \end{aligned} $

We have, in a triangle $A B C$,

$ \tan \frac{A}{2}: \tan \frac{B}{2}: \tan \frac{C}{2}=15: 10: 6 $

We know that

$ \begin{aligned} r & =\frac{\Delta}{s}=(s-a) \tan \frac{A}{2} \\\\ & =(s-b) \tan \frac{B}{2}=(s-c) \tan \frac{C}{2} \\\\ \therefore \frac{\tan \frac{A}{2}}{\tan \frac{B}{2}} & =\frac{s-b}{s-a}=\frac{15}{10}=\frac{15}{10} \times \frac{2}{2}=\frac{30}{20} \end{aligned} $

$\begin{aligned} & \text { and } \frac{\tan \frac{B}{2}}{\tan \frac{C}{2}}=\frac{s-c}{s-b}=\frac{10}{6}=\frac{10}{6} \times \frac{5}{5}=\frac{50}{30} \\\\ & \therefore \quad(s-a):(s-b):(s-c) \\\\ & \quad=20: 30: 50=2: 3: 5 \\\\ & \Rightarrow a: b: c=8: 7: 5 \\\\ & \text { Let } a=8 k, b=7 k \text { and } c=5 k \\\\ & \text { Now, } \frac{a}{b-c}=\frac{8 k}{7 k-5 k}=\frac{8 k}{2 k}=4\end{aligned}$

$ \begin{aligned} & \frac{a\left(r_1+r_2 r_3\right)}{r_1-r+r_2+r_3} \\\\ & =\frac{a\left(\frac{\Delta}{s} \cdot \frac{\Delta}{s-a}+\frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}\right)}{\left(r_1+r_2+r_3\right)-r} \\\\ & =\frac{a\left(\Delta^2 \cdot\left(\frac{(s-b)(s-c)+s(s-a)}{s(s-a)(s-b)(s-c)}\right)\right)}{(4 R-r)+r} \\\\ & =\frac{a\left(\Delta^2 \cdot \frac{s^2-(a+b+c) s+b c+s^2}{\Delta^2}\right)}{4 R} \\\\ & =\frac{a}{4 R}\left(s^2-(2 s) s+b c+s^2\right) \\\\ & =\frac{a b c}{4 R}=\Delta=\sqrt{r_1 r_2 r_3} \end{aligned} $

We have,

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

$\begin{aligned} & a^2+b^2+2 a b\left(\sin ^2 \frac{C}{2}-\cos ^2 \frac{C}{2}\right)=a^2+b^2 \\\\ & 2 a b\left(\sin ^2 \frac{C}{2}-\cos ^2 \frac{C}{2}\right)=0 \\\\ & \Rightarrow \quad \sin ^2 \frac{C}{2}-\cos ^2 \frac{C}{2}=0 \\\\ & \Rightarrow \quad \tan ^2 \frac{C}{2}=1 \\\\ & \Rightarrow \quad \tan \frac{C}{2}= \pm 1 \\\\ & \Rightarrow \quad \frac{C}{2}=45 \\\\ & \Rightarrow \quad C=90^{\circ}\end{aligned}$

So, $\cos A=\frac{b}{c} \Rightarrow \sin B=\frac{b}{c}$

$ \Rightarrow \quad \cos A=\sin B $

$\begin{aligned} & \because r_1 r_2+\pi_3=35 \\\\ & \begin{aligned} \Rightarrow & \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b}+\frac{\Delta}{s} \cdot \frac{\Delta}{s-c}=35 \\\\ \Rightarrow \quad & \Delta^2\left(\frac{s \cdot(s-c)+(s-a)(s-b)}{s(s-a)(s-b)(s-c)}\right)=35 \\\\ \Rightarrow \quad & 2 s^2-(a+b+c) s+a b=35 \\\\ & \quad\left[\because s(s-a)(s-b)(s-c)=\Delta^2\right] \\\\ \Rightarrow & 2 s^2-2 s^2+a b=35[\because a+b+c=25]\end{aligned}\end{aligned}$

$ \begin{array}{lr} \Rightarrow & a b=35 \quad \ldots \text { (i) } \\\\ \because & r_2 r_3+\pi_1=63 \\\\ \text { So, } & b c=63 \quad \ldots \text { (ii) } \end{array} $

$ \because \quad r_3 r_1+r \cdot r_2=45 $

So, $ c a=45 \quad \ldots \text { (iii) } $

From Eq. (i), (ii) and (iii), we get

$ a=5, b-7, c=9 $

Hence, $2 s=a+b+c=5+7+9=21$

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} $

The incentre of triangle formed by the points $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$, $\left(x_3, y_3, z_3\right)$ is

$ \begin{aligned} & \left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c},\frac{a z_1+b z_2+c z_3}{a+b+c}\right) \end{aligned} $

where $a, b, c$ are the sides of the triangle. The vertices are $A(0,0,0), B(3,0,0)$ and $C(0,4,0)$

Here, the sides are

$ A B=3, A C=4 $

and $B C=5$

The incentre of the triangle is

$ \begin{aligned} & \left(\frac{5(0)+4(3)+3(0)}{12}, \frac{5(0)+4(0)+3(4)}{12}, \frac{5(0)+4(0)+3(0)}{12}\right) \\ & =\left(\frac{12}{12}, \frac{12}{12}, 0\right)=(1,1,0) \end{aligned} $

We have, $r_1=4, r_2=8$ and $r_3=24$

$ \begin{aligned} \frac{1}{r} & =\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{4}+\frac{1}{8}+\frac{1}{24} \\ & =\frac{6+3+1}{24}=\frac{10}{24}=\frac{5}{12} \Rightarrow r=\frac{12}{5} \\ \Delta^2 & =r_1 \cdot r_2 \cdot r_3 \cdot r=4 \times 8 \times 24 \times \frac{12}{5} \end{aligned} $

So, $\Delta=\frac{96}{\sqrt{5}} \Rightarrow r_1=\frac{\Delta}{s-a}$

$\Rightarrow s-a=\frac{\Delta}{r_1}=\frac{96}{\sqrt{5} \times 4}=\frac{24}{\sqrt{5}}$

Similarly, $s-b=\frac{\Delta}{r_2}=\frac{96}{\sqrt{5} \times 8}=\frac{12}{\sqrt{5}} \cdots$

$ s-c=\frac{\Delta}{r_3}=\frac{96}{\sqrt{5} \times 24}=\frac{4}{\sqrt{5}} \ldots $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & s-a+s-b+s-c=\frac{24}{\sqrt{5}}+\frac{12}{\sqrt{5}}+\frac{4}{\sqrt{5}} \\ & 3 s-(a+b+c)=\frac{40}{\sqrt{5}} \\ & 3 s-2 s=\frac{40}{\sqrt{5}} \Rightarrow s=\frac{40}{\sqrt{5}} \end{aligned} $

Then, $a=s-\frac{24}{\sqrt{5}}=\frac{40}{\sqrt{5}}-\frac{24}{\sqrt{5}}=\frac{16}{\sqrt{5}}$

Let $a, b, c$ be the sides of $\triangle A B C$, tha

$ \begin{gathered} \cot A=\frac{b^2+c^2-a^2}{4 \Delta} \\ \cot B=\frac{a^2+c^2-b^2}{4 \Delta}, \cot C=\frac{a^2+b^2-c^2}{4 \Delta} \end{gathered} $

So, $\Sigma \cot A=\cot A+\cot B+\cot C=\frac{1}{45}$

$ \begin{aligned} & {\left[b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-\right.} \\ & =\frac{a^2+b^2+c^2}{4 \Delta} \end{aligned} $

(B) In $\triangle A B C$,

$ \begin{aligned} & \cot \frac{A}{2}=\frac{s(s-a)}{\sqrt{s(s-a)(s-b)(s-c)}}=\frac{s(s-c)}{\Delta} \\ & \cot \frac{B}{2}=\frac{s-b}{\Delta} \\ & \cot \frac{c}{2}=\frac{s-c}{\Delta} \text { where, } s=\frac{a+b+c}{2} \\ & \text { Thus, } \begin{aligned} & \Sigma \cot \frac{A}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{c}{2} \\ &=\frac{s(s-a)}{\Delta}+\frac{s(s-b)}{\Delta}+\frac{s(s-c)}{\Delta} \\ &=\frac{s}{\Delta}[s-a+s-b+s-c] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{s}{\Delta}[3 s-(a+b+c)] \\ & =\frac{s}{\Delta}(3 s-2 s)=\frac{s^2}{\Delta}=\frac{(a+b+c)^2}{4 \Delta} \end{aligned} $

(C) We have, $\tan A: \tan B: \tan C=1: 2: 3$

So, $\tan A=k, \tan B=2 K, \tan C=3 K$

Now, $\tan C=\tan [\pi-(A+B)]=3 K$

$\begin{aligned} & \Rightarrow \quad-\tan (A+B)=3 K \\ & \Rightarrow \quad \tan (A+B)=-3 K \\ & \Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-3 K \end{aligned}$

$ \Rightarrow \quad \frac{K+2 K}{1-2 K^2}=-3 K $

$ \Rightarrow \quad 1-2 K^2=-1 \Rightarrow 2 K^2=2 $

$ \Rightarrow \quad K^2=1 \Rightarrow K=1 $

$ \begin{aligned} & \sin A=\frac{1}{\sqrt{2}} \Rightarrow \sin B=\frac{2}{\sqrt{5}} \\ & \sin C=\frac{3}{\sqrt{10}} \end{aligned} $

$ \begin{aligned} &\begin{aligned} \sin A: \sin B & : \sin C \\ & =\frac{1}{\sqrt{2}}: \frac{2}{\sqrt{5}} \frac{: 3}{\sqrt{10}}=\sqrt{5}: 2 \sqrt{2}: 3 \end{aligned}\\ &\text { (D) We have, }\\ &\begin{aligned} & \cot \left(\frac{A}{2}\right): \cos \left(\frac{B}{2}\right): \cot \left(\frac{C}{2}\right)=3: 5: 7 \\ \Rightarrow & \frac{\cot A / 2}{3}=\frac{\cot B / 2}{7}=\frac{\cot C / 2}{9}=k \\ \Rightarrow \quad & \frac{s(s-a)}{3 \Delta}=\frac{s(s-b)}{7 \Delta}=\frac{s(s-c)}{9 \Delta} \\ \Rightarrow \quad & \frac{s-a}{3}=\frac{s-b}{7}=\frac{s-c}{9} \\ \Rightarrow \quad & \frac{s-b+s-c}{9+7}=\frac{s-a+s-c}{3+9} \\ \Rightarrow & =\frac{s-a+s-b}{3+7} \\ \Rightarrow & \frac{2 s-b}{16}=\frac{a}{16}=\frac{b}{12}=\frac{c}{10}=K \\ \Rightarrow & \quad a: b: c=16: 12: 10=8: 6: 5 \end{aligned} \end{aligned} $

$\begin{aligned} & r_1=2 r_2=3 r_3 \\ & \Rightarrow \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c} \\ & \Rightarrow \frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=k \text { (let) } \\ & \Rightarrow s-a=\frac{1}{k}, s-b=\frac{2}{k} \text { and } s-c=\frac{3}{k}, s=\frac{6}{k} \\ & \Rightarrow a=s-(s-a)=\frac{5}{k} \\ & \text { Similarly, } b=\frac{4}{k} \text { and } c=\frac{3}{k} \\ & \text { We know that } \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \Rightarrow \frac{\sin A}{5}=\frac{\sin B}{4}=\frac{\sin C}{3} \\ & \Rightarrow \sin A: \sin B: \sin C=5: 4: 3\end{aligned}$

The circumradius $ R $ for a right-angled triangle is given by:

$ \frac{b}{\sin B} = 2R $

Since $\sin 90^\circ = 1$, we have:

$ b = 2R $

Therefore, $ R = \frac{b}{2} $.

The inradius $ r $ is given by:

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

Since $\angle B = 90^\circ$, $\frac{B}{2} = 45^\circ$ and $\tan 45^\circ = 1$. Hence:

$ r = s-b $

Now, we compute $ 2(r + R) $:

$ 2(r + R) = 2 \left(s - b + \frac{b}{2}\right) = 2s - b $

Further simplifying, considering the semi-perimeter $ s = \frac{a+b+c}{2} $:

$ 2s = a + b + c $

Therefore, substituting back:

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

So, the value is $ a + c $.

In a triangle $ \triangle ABC $, we have the relation:

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

Rearranging the terms, we obtain:

$ (s-b)(s-c) + (s-b)(s-a) = 2(s-a)(s-c) $

Next, divide each term by $ (s-a)(s-b)(s-c) $ and multiply by the area of the triangle $ \Delta $:

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

This simplifies to:

$ r_1 + r_3 = 2r_2 $

We have,

ABC is a triangle

$ \cos A+\cos B+\cos C $

$ \begin{aligned} & =2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{C}{2} \\ & =1+2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)-2 \sin ^2 \frac{C}{2} & {[\because A+B+C=\pi]} \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right] \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right] \\ & =1+2 \sin \frac{C}{2}\left[2 \sin \frac{A}{2} \sin \frac{B}{2}\right] \\ & =1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{aligned} $

$ \begin{aligned} & =1+4\left[\sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}}\right] \\ & =\frac{1+4(s-a)(s-b)(s-c)}{a b c} \\ & =\frac{1+4 s(s-a)(s-b)(s-c)}{s a b c} [\therefore \text { multiply and divide by } s] \\ & =1+4 \frac{\Delta^2}{s a b c}, \text { where } \Delta \text { is area of triangle } \\ & =1+\left(\frac{\Delta}{s}\right)\left(\frac{4 \Delta}{a b c}\right) \quad\left[\because r=\frac{\Delta}{s} \text { and } R=\frac{a b c}{4 \Delta}\right] \\ & =1+\frac{r}{R} \end{aligned} $

In triangle $ \triangle ABC $, given $ a = 26 $, $ b = 30 $, and $ \cos C = \frac{63}{65} $, we need to find the length of side $ c $.

We know the cosine rule for any triangle is given by:

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

Substituting the known values:

$ \frac{a^2 + b^2 - c^2}{2 \times a \times b} = \frac{63}{65} $

This implied expression becomes:

$ \frac{26^2 + 30^2 - c^2}{2 \times 26 \times 30} = \frac{63}{65} $

Now, solve this step-by-step:

Calculate $ 26^2 $ and $ 30^2 $:

$ 26^2 = 676, \quad 30^2 = 900 $

Substitute these into the equation:

$ \frac{676 + 900 - c^2}{1560} = \frac{63}{65} $

Simplify the numerator:

$ 676 + 900 = 1576 $

Substitute back:

$ \frac{1576 - c^2}{1560} = \frac{63}{65} $

Cross-multiply to solve for $ c^2 $:

$ 65(1576 - c^2) = 63 \times 1560 $

Compute $ 63 \times 1560 $:

$ 63 \times 1560 = 98280 $

Now, solve for $ c^2 $:

$ 65 \times (1576 - c^2) = 98280 $

Simplify the expressions:

$ 102440 - 65c^2 = 98280 $

Rearrange to find $ c^2 $:

$ 102440 - 98280 = 65c^2 $

Solve:

$ 4160 = 65c^2 $

Divide by 65 to isolate $ c^2 $:

$ c^2 = \frac{4160}{65} = 64 $

Take the square root:

$ c = \sqrt{64} = 8 $

Thus, the value of $ c $ is 8.

In $\triangle AFC$

$ \cos A=\frac{A F}{A C}=\frac{A F}{b} $

$ \Rightarrow \quad A F=b \cos A $

In $\triangle AFH$

$ \cos \left(90^{\circ}-2 B=\frac{A F}{A H} \Rightarrow \sin B=\frac{b \cos A}{x}\right. $

$ \Rightarrow \quad x=\left(\frac{b}{\sin B}\right) \cos A $

Using sine rule, we get

$ \begin{aligned} x & =\left(\frac{a}{\sin A}\right) \cos A \\ \Rightarrow \quad & \frac{a}{x}=\tan A \end{aligned}$

Similarly, $\frac{b}{y}=\tan B$ and $\frac{c}{z}=\tan c$

$ \begin{aligned} \therefore \quad \frac{a k c}{x y z} & =\tan A \tan B \tan C \\ & =\tan A+\tan B+\tan C \quad [\because A+B+C=\pi]\\ & =\frac{a}{x}+\frac{b}{y}+\frac{c}{z} \end{aligned} $

$ \begin{aligned} &\text { Given, } a, b \text { and } c \text { are in AP. }\\ &\begin{aligned} & \text { Also, } A=2 C \Rightarrow 2 C+B+C=\pi \\ & \Rightarrow B=\pi-3 C \Rightarrow 2 b=a+c \\ & 2 \sin B=\sin A+\sin C \text { [from sine rule] } \\ & \Rightarrow 2 \sin B=\sin 2 C+\sin C \\ & \Rightarrow 2 \sin (\pi-3 C)=2 \sin C \cos C+\sin C \\ & \Rightarrow 2\left(3 \sin C-4 \sin ^3 C\right)=\sin C(2 \cos C+1) \\ & \Rightarrow \quad 6-8 \sin ^2 C=2 \cos C+1 \\ & \Rightarrow 6-8\left(1-\cos ^2 C\right)=+2 \cos C+1 \\ & \Rightarrow 8 \cos ^2 C-2 \cos C-3=0 \\ & \Rightarrow \cos C=\frac{3}{4}, \frac{-1}{2} \end{aligned} \end{aligned} $

$ \text { Now, } \begin{aligned} \frac{b}{c} & =\frac{\sin B}{\sin C}=\frac{\sin 3 C}{\sin C}=3-4 \sin ^2 C \\ & =4 \cos ^2 C-1 \end{aligned} $

$ \begin{aligned} &\text { At }-\frac{1}{2}, b: c=0\\ &\text { Now, at } \frac{3}{4}, b: c=5: 4 \end{aligned} $

Since, in an equilateral triangle, the in-radius, circumradius and one of the ex-radii are in the ratio i.e.,

$ r: R: r_2=1: 2: 3 $

So, here, both Assertion (A) and Reason (R) both are false.

If $a: b: c=4: 5: 6$

$ \begin{aligned} a=4 k, b & =5 k, c=6 k, s=\frac{15 k}{2} \\ (s-a) & =\frac{7 k}{2},(s-b)=\frac{5 k}{2} \text { and }(s-c)=\frac{3 k}{2} \\ \Delta^2 & =s \cdot(s-a) \cdot(s-b) \cdot(s-c) \\ & =\frac{15 k}{2} \times \frac{7 k}{2} \times \frac{5 k}{2} \times \frac{3 k}{2} \\ \Delta & =15 \sqrt{7}\left(\frac{k}{2}\right)^2 \\ r & =\frac{\Delta}{s}=\frac{15 \sqrt{7}\left(\frac{k}{2}\right)^2}{\frac{15 k}{2}}=\sqrt{7} \cdot \frac{k}{2} \\ R & =\frac{a b c}{4 \Delta}=\frac{4 \times 5 \times 6 \times k^3}{4 \times 15 \sqrt{7}\left(\frac{k}{2}\right)^2}=\frac{8}{\sqrt{7}} k \\ \frac{R}{r} & =\frac{\frac{8}{\sqrt{7}} \cdot k}{\frac{\sqrt{7}}{2} \cdot k}=\frac{16}{7} \end{aligned} $

Hence, the answer is $16: 7$.

We have,

$ \begin{aligned} & a+b+c=\frac{6(\sin A+\sin B+\sin C)}{3} \\ & k(\sin A+\sin B+\sin C) \\ & =2(\sin A+\sin B+\sin C) \\ & \text { where, } k=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & k=2 \quad(\because \sin A+\sin B+\sin C \neq 0) \\ & \frac{a}{\sin A}=2 \Rightarrow \sin A=\frac{1}{2} \end{aligned} $

Then, $\angle A=\frac{\pi}{6}$

Given, $b=6, c=7$ and $\tan \frac{A}{2}=\frac{1}{\sqrt{6}}$

$ \begin{aligned} & \because \quad \tan \frac{A}{2}=\frac{1}{\sqrt{6}} \text { and } \cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}} \\ & \therefore \quad \cos A=\frac{1-\left(\frac{1}{\sqrt{6}}\right)^2}{1+\left(\frac{1}{\sqrt{6}}\right)^2}=\frac{5 / 6}{7 / 6}=\frac{5}{7} \end{aligned} $

Now, by using the cosine formula,

$ \begin{aligned} & a^2=b^2+c^2-2 b c \cos A \\ \Rightarrow \quad & a^2=6^2+7^2-2(6)(7)\left(\frac{5}{7}\right) \\ \Rightarrow & a^2=25 \Rightarrow a=5\,\,\,\,\,\,[\because a>0] \end{aligned} $

Now, area of $\triangle A B C=\frac{1}{2} b c \sin A$

$ \begin{aligned} & =\frac{1}{2} \times 6 \times 7 \times \sqrt{1-\left(\frac{5}{7}\right)^2} \\ & =6 \sqrt{6} \\ \text { Perimeter } & =a+b+c=5+6+7=18 \\ \therefore \text { Inradius } & =\frac{2 \times \text { Area of } \triangle A B C}{\text { Perimeter }} \\ & =\frac{2 \times 6 \sqrt{6}}{18}=\frac{2 \sqrt{6}}{3} \end{aligned} $

Given, $a=7, b=8$ and $c=9$

∴ Semi-perimeter

$ \begin{aligned} & =s=\frac{a+b+c}{2}=\frac{7+8+9}{2}=12 \\ \text { Area } & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{12(12-7)(12-8)(12-9)} \\ & =\sqrt{12 \times 5 \times 4 \times 3} \\ & =12 \sqrt{5} \\ \text { Now, } r_1 & =\frac{\text { Area }}{s-a}=\frac{12 \sqrt{5}}{12-7}=\frac{12 \sqrt{5}}{5}=\frac{12}{5} \sqrt{5} \\ r_2 & =\frac{\text { Area }}{s-b}=\frac{12 \sqrt{5}}{12-8}=\frac{12 \sqrt{5}}{4}=\frac{12}{4} \sqrt{5} \\ r_3 & =\frac{\text { Area }}{s-c}=\frac{12 \sqrt{5}}{12-9}=\frac{12 \sqrt{5}}{3}=\frac{12}{3} \sqrt{5} \end{aligned} $

$ \begin{aligned} & \therefore \frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\left(\frac{5}{12 \sqrt{5}}\right)^2 +\left(\frac{4}{12 \sqrt{5}}\right)^2+\left(\frac{3}{12 \sqrt{5}}\right)^2 \\ & \quad=\frac{25+16+9}{12 \times 12 \times 5}=\frac{5 \times 5 \times 2}{12 \times 12 \times 5}=\frac{5}{72} \end{aligned} $

$b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$

$A$ is acute angle

$ \begin{aligned} \cos A & =\sqrt{1-\sin ^2 A}=\sqrt{1-\frac{56}{81}}=\frac{5}{9} \\ \therefore \quad a^2 & =b^2+c^2-2 b c \cos A \\ & =36+81-2 \times 6 \times 9 \times \frac{5}{9}=57 \end{aligned} $

$ \begin{aligned}Now\,\, & \cos B+\cos C=\frac{c^2+a^2-b^2}{2 a c}+\frac{a^2+b^2-c^2}{2 a b} \\ & \Rightarrow a(\cos B+\cos C)=\frac{c^2+a^2-b^2}{2 c} \\ & +\frac{a^2+b^2-c^2}{2 b} \\ & =\frac{81+57-36}{2 \times 9}+\frac{57+36-81}{2 \times 6} \\ & \Rightarrow 3 a(\cos B+\cos C)=\frac{102}{6}+\frac{12}{4}=20 \end{aligned} $

Given, equation

$ x^3-11 x^2+36 x-36=0 $

Let the roots are $\alpha, \beta$ and $\gamma$.

If $\alpha, \beta$ and $\gamma$ are ex-radii of triangle, then

$ S^2=\alpha \beta+\beta \gamma+\gamma \alpha $

where $S=$ semi-perimeter

$ \begin{aligned} & \Rightarrow \quad S^2=36 \Rightarrow S=6 \\ & \text { Perimeter }=6 \times 2=12 \end{aligned} $

Given, $P Q R$ is an isosceles triangle, with $P Q=Q R$. and also the radius of circumcircle of $\triangle P Q R$ is equal to the length of $P Q$,

So, $P Q=P R=r$ (say) and $\angle Q=\angle R$

So, sign rule $\frac{P Q}{\sin \angle R}=2 r$

$ \begin{aligned} & \Rightarrow \frac{n}{\sin \angle R}=2 r \Rightarrow \sin \angle R=\frac{1}{2}=\sin \frac{\pi}{6} \\ & \therefore \angle R=\frac{\pi}{6} \end{aligned} $

Now, $\angle P=\pi-(\angle Q+\angle R)=\pi-\left(\frac{\pi}{6}+\frac{\pi}{6}\right)$

$ \begin{aligned} & =\pi-\frac{\pi}{3} \\ & =\frac{2 \pi}{3}=\frac{2 \times 180^{\circ}}{3}=120^{\circ} \end{aligned} $

Given, $\triangle A B C$ and

$ \frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c} \text { (given) } $

We know that, $a=2 R \sin A, b=2 R \sin B$, and $c=2 R \sin C$

$ \begin{aligned} & \Rightarrow \frac{\cos A}{2 R \sin A}=\frac{\cos B}{2 R \sin B}=\frac{\cos C}{2 R \sin C} \\ & =\tan A=\tan B=\tan C \end{aligned} $

So, this know that the $\triangle A B C$ is an equilateral triangle.

Thus, $a=b=c=2$ (obviously)

$ \begin{aligned} \therefore \text { Area } & =\frac{\sqrt{3}}{4}(a)^2=\frac{\sqrt{3}}{4} \times(2)^2 \\ & =\sqrt{3} \text { sq units } \end{aligned} $

$\theta+\phi=45^{\circ}$...(i)

In $\triangle B C D$,

$ \begin{aligned} \tan \theta & =\frac{C D}{B C}=\frac{\frac{x}{2}}{x}=\frac{1}{2} \\ \Rightarrow \quad \cot \theta & =2 \\ \phi & =45^{\circ}-\theta \quad \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad \tan \phi & =\tan \left(45^{\circ}-\theta\right) \\ & =\frac{\tan 45^{\circ}-\tan \theta}{1+\tan 45^{\circ} \cdot \tan \theta} \\ & =\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3} \\ \Rightarrow \quad \cot \phi & =3 \end{aligned} $

$\therefore$ Possible values of the cotangents of the two angles are 2 and 3.

Given, $r_1=2 r_2=3 r_3$

$ \begin{aligned} & \text { Let } r_1=2 r_2=3 r_3=k \\ & r_1=k \\ & \Rightarrow \quad \frac{\Delta}{s-a}=k \Rightarrow s-a=\frac{\Delta}{k} \\ & a=s-\frac{\Delta}{k} \\ & 2 r_2=k \\ & \Rightarrow \frac{2 \Delta}{s-b}=k \Rightarrow s-b=\frac{2 \Delta}{k} \\ & \Rightarrow \quad b=s-\frac{2 \Delta}{k} \\ & \Rightarrow \quad 3 r_3=k \\ & \frac{3 \Delta}{s-c}=k \Rightarrow s-c=\frac{3 \Delta}{k} \end{aligned} $

$ \Rightarrow \quad c=s-\frac{3 \Delta}{k} $

Adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} a+b+c & =3 s-\frac{6 \Delta}{k} \\ \Rightarrow \quad 2 s & =3 s-\frac{6 \Delta}{k} \Rightarrow s=\frac{6 \Delta}{k} \\ a & =\frac{5 \Delta}{k}, b=\frac{4 \Delta}{k} \text { and } c=\frac{3 \Delta}{k} \\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{5}{4}+\frac{4}{3}+\frac{3}{5} \\ & =\frac{75+80+36}{60}=\frac{191}{60} \end{aligned} $

$ \begin{aligned} & \text { } a=7, b=8 \text { and } \tan C=\frac{3 \sqrt{5}}{2} \\ & \qquad \begin{array}{ll} & \tan C=\frac{3 \sqrt{5}}{2}=\frac{P}{B} \\ \Rightarrow & H=7 \\ \therefore & \cos C=\frac{B}{H}=\frac{2}{7} \\ \because & \cos C=\frac{a^2+b^2-c^2}{2 a b} \\ & c^2=a^2+b^2-2 a b \cos c \\ & c^2=49+64-2 \cdot 7 \cdot 8\left(\frac{2}{7}\right) \\ & c^2=113-32=81 \\ \Rightarrow & c=9 \end{array} \end{aligned} $

$ \frac{a}{\tan A}=\frac{b}{\tan B}=\frac{c}{\tan C} $

$ \Rightarrow \frac{a}{\sin A} \cdot \cos A=\frac{b}{\sin B} \cdot \cos B=\frac{c}{\sin C} \cdot \cos C \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

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

∴ Eq. (i) becomes,

$ \begin{aligned} & & 2 R \cos A & =2 R \cos B=2 R \cos C\\ \Rightarrow & & \cos A & =\cos B=\cos C \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

or $\quad A=B=C$      [i.e. $60^{\circ}$ ]

$ \begin{aligned} \therefore \cos ^2 A+\cos ^2 B+ & \cos ^2 C \\ = & 3 \cos ^2 A=3 \cos ^2 60^{\circ} \\ = & 3 \times\left(\frac{1}{2}\right)^2=\frac{3}{4} \end{aligned} $

In $\triangle A B C, a=7, b=10$ and $c=11$

$ \begin{aligned} R & =\frac{a b c}{4 \Delta} \text { and } r=\frac{\Delta}{S} \\ \therefore \quad \frac{R}{r} & =\frac{\left(\frac{a b c}{4 \Delta}\right)}{\left(\frac{\Delta}{S}\right)} \\ \frac{a b c}{4 \Delta^2} \times S &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ S & =\frac{a+b+c}{2} \\ & =\frac{7+10+11}{2}=\frac{28}{2}=14 \\ \Delta & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{14(7)(4)(3)} \\ \Rightarrow \quad \Delta^2 & =14 \times 7 \times 4 \times 3 \end{aligned} $

From Eq. (i), we have

$ \begin{aligned} \frac{R}{r} & =\frac{7 \times 10 \times 11}{4 \times 14 \times 7 \times 4 \times 3} \times 14 \\ & =\frac{5 \times 11}{2 \times 4 \times 3}=\frac{55}{24} \end{aligned} $