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} $

$ 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)} $

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

$ \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} $

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

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

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

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

We know by angle sum property,

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

Using sine rule,

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

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

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

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

In $\triangle A B D$,

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

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

In $\triangle A G B$,

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

As the median divides the triangle in equal parts

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

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

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

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

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

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

Given, perimeter $=42$

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

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

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

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

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

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

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

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