Relation between sides and angles, Solutions of triangles Questions in English

(D) Given that $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in arithmetic progression,we have $\frac{2}{r_2} = \frac{1}{r_1} + \frac{1}{r_3}$.
Using the formulas $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$ and $r = \frac{\Delta}{s}$,we get:
$\frac{2(s-b)}{\Delta} = \frac{s-a}{\Delta} + \frac{s-c}{\Delta}$
$2s - 2b = 2s - (a+c)$
$2b = a+c$.
This implies that $a, b, c$ are in arithmetic progression.
Now,we need to find $r_2 : r = \frac{r_2}{r} = \frac{\Delta / (s-b)}{\Delta / s} = \frac{s}{s-b}$.
Since $2b = a+c$,we have $s = \frac{a+b+c}{2} = \frac{3b}{2}$.
Thus,$\frac{s}{s-b} = \frac{3b/2}{3b/2 - b} = \frac{3b/2}{b/2} = 3$.
Therefore,$r_2 : r = 3 : 1$.

(A) Given $r_1 > r_2 > r_3$.
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these values,we get $\frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}$.
Since $\Delta > 0$,we have $s-a < s-b < s-c$.
Subtracting $s$ from all parts,we get $-a < -b < -c$.
Multiplying by $-1$ reverses the inequality signs,resulting in $a > b > c$.

(C) Given,in $\triangle ABC$,$r_1=12, r_2=18, r_3=36$.
We know that $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$.
Substituting the values: $\frac{1}{r} = \frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{3+2+1}{36} = \frac{6}{36} = \frac{1}{6}$.
Thus,$r = 6$.
Also,$\Delta^2 = r_1 r_2 r_3 r = 12 \times 18 \times 36 \times 6 = 46656$.
So,$\Delta = \sqrt{46656} = 216$.
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r} = \frac{216}{6} = 36$.
We know $r_2 = \frac{\Delta}{s-b}$.
Substituting the values: $18 = \frac{216}{36-b}$.
$36-b = \frac{216}{18} = 12$.
$b = 36 - 12 = 24$.

(A) Given: $\left(1-\frac{r_1}{r_2}\right)\left(1-\frac{r_1}{r_3}\right)=2$
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these values:
$\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right) = 2$
$\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right) = 2$
$\left(\frac{s-a-s+b}{s-a}\right)\left(\frac{s-a-s+c}{s-a}\right) = 2$
$\left(\frac{b-a}{s-a}\right)\left(\frac{c-a}{s-a}\right) = 2$
$(b-a)(c-a) = 2(s-a)^2$
$bc - ab - ac + a^2 = 2\left(\frac{b+c-a}{2}\right)^2$
$bc - ab - ac + a^2 = 2\frac{(b+c-a)^2}{4}$
$2(bc - ab - ac + a^2) = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$
$2bc - 2ab - 2ac + 2a^2 = b^2 + c^2 + a^2 + 2bc - 2ab - 2ac$
$2a^2 = b^2 + c^2 + a^2$
$a^2 = b^2 + c^2$
Since $a^2 = b^2 + c^2$,the triangle is a right-angled triangle.

(C) Given $A = 30^{\circ} + C$ and $R = (\sqrt{3} + 1)r$.
We know that $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
Substituting $r = \frac{R}{\sqrt{3} + 1}$,we get $\frac{1}{\sqrt{3} + 1} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$.
Rationalizing the denominator,$\frac{\sqrt{3} - 1}{2} = 4 \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,so $\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) = \frac{\sqrt{3} - 1}{8}$.
Using $A + B + C = 180^{\circ}$,we have $B = 180^{\circ} - (A + C) = 180^{\circ} - (30^{\circ} + 2C) = 150^{\circ} - 2C$.
Substituting these values into the identity and solving for the angles,we find $\angle A = 75^{\circ}, \angle B = 60^{\circ}, \angle C = 45^{\circ}$.
Since $\angle B = 60^{\circ}$ and $\angle A + \angle C = 120^{\circ}$,the triangle satisfies the given conditions.
Thus,the triangle is not right-angled,equilateral,or acute-angled (since $75^{\circ} > 90^{\circ}$ is false,but $75^{\circ}$ is acute,$60^{\circ}$ is acute,$45^{\circ}$ is acute).
Wait,$75^{\circ}, 60^{\circ}, 45^{\circ}$ are all acute,so $ABC$ is acute-angled.

(C) Let the sides $a, b, c$ be in $A$.$P$.,so $2b = a + c$.
We know that the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
To check if $r_1, r_2, r_3$ are in $H$.$P$.,we check if $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in $A$.$P$.
$\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,$\frac{1}{r_3} = \frac{s-c}{\Delta}$.
Since $a, b, c$ are in $A$.$P$.,$s-a, s-b, s-c$ are also in $A$.$P$. because $s-a + s-c = 2s - (a+c) = 2s - 2b = 2(s-b)$.
Thus,$\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$ are in $A$.$P$.,which implies $r_1, r_2, r_3$ are in $H$.$P$.

(A) In a right-angled triangle $ABC$ with $\angle A = 90^{\circ}$,the inradius $r$ is given by $r = \frac{b+c-a}{2}$.
The circumradius $R$ is given by $R = \frac{a}{2}$.
Therefore,$r+R = \frac{b+c-a}{2} + \frac{a}{2} = \frac{b+c}{2}$.
Multiplying by $2$,we get $2(r+R) = b+c$.

(A) We know that $r = \frac{\Delta}{s}$,$r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Substituting these into the expression:
$\frac{r_1-r}{a} + \frac{r_2-r}{b} + \frac{r_3-r}{c} = \frac{1}{a}\left(\frac{\Delta}{s-a} - \frac{\Delta}{s}\right) + \frac{1}{b}\left(\frac{\Delta}{s-b} - \frac{\Delta}{s}\right) + \frac{1}{c}\left(\frac{\Delta}{s-c} - \frac{\Delta}{s}\right)$
$= \frac{1}{a}\left(\frac{\Delta s - \Delta(s-a)}{s(s-a)}\right) + \frac{1}{b}\left(\frac{\Delta s - \Delta(s-b)}{s(s-b)}\right) + \frac{1}{c}\left(\frac{\Delta s - \Delta(s-c)}{s(s-c)}\right)$
$= \frac{1}{a}\left(\frac{\Delta a}{s(s-a)}\right) + \frac{1}{b}\left(\frac{\Delta b}{s(s-b)}\right) + \frac{1}{c}\left(\frac{\Delta c}{s(s-c)}\right)$
$= \frac{\Delta}{s(s-a)} + \frac{\Delta}{s(s-b)} + \frac{\Delta}{s(s-c)}$
$= \frac{r_1}{s} + \frac{r_2}{s} + \frac{r_3}{s} = \frac{r_1+r_2+r_3}{s}$.

(B) We know that $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$.
Substituting this into the expression,we get:
$a \cdot \frac{s(s-a)}{bc} + b \cdot \frac{s(s-b)}{ac} + c \cdot \frac{s(s-c)}{ab}$
$= \frac{s}{abc} [a^2(s-a) + b^2(s-b) + c^2(s-c)]$
$= \frac{s}{abc} [s(a^2+b^2+c^2) - (a^3+b^3+c^3)]$
Using the identity $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$ and $s = \frac{a+b+c}{2}$,we simplify the expression to $s + \frac{\Delta}{R}$.

(C) Given $A = 180^{\circ} - (B + C) = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
$a = 2(\sqrt{3}+1)$,$\sin 75^{\circ} = \sin(45^{\circ}+30^{\circ}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
So,$\frac{2(\sqrt{3}+1)}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = 4\sqrt{2}$.
Thus,$b = 4\sqrt{2} \sin 45^{\circ} = 4\sqrt{2} \times \frac{1}{\sqrt{2}} = 4$.
$c = 4\sqrt{2} \sin 60^{\circ} = 4\sqrt{2} \times \frac{\sqrt{3}}{2} = 2\sqrt{6}$.
Area $= \frac{1}{2} ab \sin C = \frac{1}{2} \times 2(\sqrt{3}+1) \times 4 \times \sin 60^{\circ} = 4(\sqrt{3}+1) \times \frac{\sqrt{3}}{2} = 2(3+\sqrt{3}) = 6+2\sqrt{3}$.

(B) We know that $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$.
Given $r_1=8, r_2=12, r_3=24$,we have $\frac{1}{r} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3+2+1}{24} = \frac{6}{24} = \frac{1}{4}$.
Thus,$r=4$.
Also,$r_1 = \frac{\Delta}{s-a} \implies 8 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b} \implies 12 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c} \implies 24 = \frac{\Delta}{s-c}$.
Since $\Delta = rs = 4s$,we have $s-a = \frac{4s}{8} = \frac{s}{2} \implies a = \frac{s}{2}$.
$s-b = \frac{4s}{12} = \frac{s}{3} \implies b = \frac{2s}{3}$.
$s-c = \frac{4s}{24} = \frac{s}{6} \implies c = \frac{5s}{6}$.
Using $s = \frac{a+b+c}{2} = \frac{1}{2}(\frac{s}{2} + \frac{2s}{3} + \frac{5s}{6}) = \frac{1}{2}(\frac{3s+4s+5s}{6}) = s$,which is consistent.
Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 4s$,we get $\sqrt{s(\frac{s}{2})(\frac{s}{3})(\frac{s}{6})} = 4s$.
$\sqrt{\frac{s^4}{36}} = 4s \implies \frac{s^2}{6} = 4s \implies s = 24$.
Then $a = \frac{24}{2} = 12$,$b = \frac{2(24)}{3} = 16$,$c = \frac{5(24)}{6} = 20$.
So,$(a, b, c) = (12, 16, 20)$.

(A) We know that the exradii of a triangle are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Given $2r_1 = 3r_2 = r_3 = k$ (let).
Then $r_1 = \frac{k}{2}$,$r_2 = \frac{k}{3}$,and $r_3 = k$.
Thus,$\frac{1}{r_1} = \frac{2}{k}$,$\frac{1}{r_2} = \frac{3}{k}$,and $\frac{1}{r_3} = \frac{1}{k}$.
We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}$,$\frac{1}{r_2} = \frac{s-b}{\Delta}$,and $\frac{1}{r_3} = \frac{s-c}{\Delta}$.
Summing these,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$.
Also,$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{2}{k} + \frac{3}{k} + \frac{1}{k} = \frac{6}{k} = \frac{1}{r}$,so $r = \frac{k}{6}$.
Using $s-a = \frac{\Delta}{r_1} = \frac{\Delta}{k/2} = \frac{2\Delta}{k}$,$s-b = \frac{3\Delta}{k}$,and $s-c = \frac{\Delta}{k}$.
Since $s = \frac{6\Delta}{k}$,we have $a = s - (s-a) = \frac{6\Delta}{k} - \frac{2\Delta}{k} = \frac{4\Delta}{k}$,$b = \frac{3\Delta}{k}$,and $c = \frac{5\Delta}{k}$.
Therefore,$a : b : c = 4 : 3 : 5$.

(D) The area of $\triangle ABC$ is given by $\Delta = \frac{1}{2} ac \sin B$.
Given $\Delta = 6+2\sqrt{3}$,$a = 2(\sqrt{3}+1)$,and $B = 45^{\circ}$.
Substituting these values: $6+2\sqrt{3} = \frac{1}{2} \times 2(\sqrt{3}+1) \times c \times \sin 45^{\circ}$.
$6+2\sqrt{3} = (\sqrt{3}+1) \times c \times \frac{1}{\sqrt{2}}$.
$c = \frac{\sqrt{2}(6+2\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}(3+\sqrt{3})}{\sqrt{3}+1} = \frac{2\sqrt{2}\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = 2\sqrt{6}$.
Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.
$b^2 = [2(\sqrt{3}+1)]^2 + (2\sqrt{6})^2 - 2[2(\sqrt{3}+1)](2\sqrt{6}) \cos 45^{\circ}$.
$b^2 = 4(3+1+2\sqrt{3}) + 24 - 8\sqrt{6}(\sqrt{3}+1) \times \frac{1}{\sqrt{2}}$.
$b^2 = 16 + 8\sqrt{3} + 24 - 8\sqrt{3}(\sqrt{3}+1) = 40 + 8\sqrt{3} - 24 - 8\sqrt{3} = 16$.
Therefore,$b = \sqrt{16} = 4$.

(C) In $\triangle ABC$,the sum of angles is $A + B + C = 180^{\circ}$.
$A + B = 180^{\circ} - C$
Taking $\cot$ on both sides:
$\cot(A + B) = \cot(180^{\circ} - C)$
Using the formula $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(180^{\circ} - C) = -\cot C$:
$\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$
$\cot A \cot B - 1 = -\cot C(\cot A + \cot B)$
$\cot A \cot B - 1 = -\cot C \cot A - \cot C \cot B$
Rearranging the terms:
$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$

(B) Given: $\cot \left(\frac{A}{2}\right)=\sqrt{\frac{1+a}{1-a}} \cot \left(\frac{\theta}{2}\right)$
Squaring both sides: $\cot^2 \left(\frac{A}{2}\right) = \left(\frac{1+a}{1-a}\right) \cot^2 \left(\frac{\theta}{2}\right)$
Using $\cot^2 \left(\frac{x}{2}\right) = \frac{1+\cos x}{1-\cos x}$,we get:
$\frac{1+\cos A}{1-\cos A} = \left(\frac{1+a}{1-a}\right) \frac{1+\cos \theta}{1-\cos \theta}$
$\frac{1+\cos \theta}{1-\cos \theta} = \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)} = \frac{1-a+\cos A-a \cos A}{1+a-\cos A-a \cos A}$
Applying Componendo and Dividendo:
$\frac{(1+\cos \theta)+(1-\cos \theta)}{(1+\cos \theta)-(1-\cos \theta)} = \frac{(1-a+\cos A-a \cos A)+(1+a-\cos A-a \cos A)}{(1-a+\cos A-a \cos A)-(1+a-\cos A-a \cos A)}$
$\frac{2}{2 \cos \theta} = \frac{2-2a \cos A}{2 \cos A-2a}$
$\frac{1}{\cos \theta} = \frac{1-a \cos A}{\cos A-a}$
Therefore,$\cos \theta = \frac{\cos A-a}{1-a \cos A}$

(A) The sides of the triangle are $a=3, b=5, c=7$.
First,we find the cosine of angle $C$ using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{3^2 + 5^2 - 7^2}{2(3)(5)} = \frac{9 + 25 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,we have $C = 120^\circ$.
The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}(3)(5) \sin(120^\circ) = \frac{15}{2} \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{4}$.
The circumradius $R$ is given by $R = \frac{abc}{4\Delta} = \frac{3 \times 5 \times 7}{4 \times (\frac{15\sqrt{3}}{4})} = \frac{105}{15\sqrt{3}} = \frac{7}{\sqrt{3}}$.

(C) Given equations: $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$.
Using the identities $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and $r = \frac{\Delta}{s}$,we know that $r_1 r_2 + r_3 r = ab$,$r_2 r_3 + r r_1 = bc$,and $r_3 r_1 + r r_2 = ac$.
Thus,$ab = 35$,$bc = 63$,and $ac = 45$.
Multiplying these gives $(abc)^2 = 35 \times 63 \times 45 = 99225$,so $abc = 315$.
Then $c = \frac{abc}{ab} = \frac{315}{35} = 9$,$a = \frac{abc}{bc} = \frac{315}{63} = 5$,and $b = \frac{abc}{ac} = \frac{315}{45} = 7$.
Therefore,$2s = a + b + c = 5 + 7 + 9 = 21$.

(A) Using the Law of Cosines:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3^2 + 7^2 - 5^2}{2 \times 3 \times 7} = \frac{9 + 49 - 25}{42} = \frac{33}{42} = \frac{11}{14}$
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{5^2 + 7^2 - 3^2}{2 \times 5 \times 7} = \frac{25 + 49 - 9}{70} = \frac{65}{70} = \frac{13}{14}$
Using the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$,so $\sin A = ak, \sin B = bk, \sin C = ck$.
Since $\sin(A+B) = \sin(180^\circ - C) = \sin C$,we have:
$\frac{\sin(A-B)}{\sin(A+B)} = \frac{\sin A \cos B - \cos A \sin B}{\sin C} = \frac{ak \cos B - \cos A bk}{ck} = \frac{a \cos B - b \cos A}{c}$
Substituting the values:
$= \frac{5 \times \frac{13}{14} - 3 \times \frac{11}{14}}{7} = \frac{\frac{65-33}{14}}{7} = \frac{32}{14 \times 7} = \frac{32}{98} = \frac{16}{49}$
Therefore,$\sqrt{\frac{\sin(A-B)}{\sin(A+B)}} = \sqrt{\frac{16}{49}} = \frac{4}{7}$.

(B) Given $b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$.
Since $A$ is an acute angle,$\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-\frac{56}{81}} = \frac{5}{9}$.
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.
$\frac{5}{9} = \frac{36+81-a^2}{2(6)(9)}$ $\Rightarrow \frac{5}{9} = \frac{117-a^2}{108}$ $\Rightarrow 60 = 117-a^2$ $\Rightarrow a^2 = 57$.
Now,$3a(\cos B+\cos C) = 3a\left(\frac{a^2+c^2-b^2}{2ac} + \frac{a^2+b^2-c^2}{2ab}\right)$.
$= 3a\left(\frac{b(a^2+c^2-b^2) + c(a^2+b^2-c^2)}{2abc}\right) = \frac{3}{2bc} [a^2(b+c) + bc^2 - b^3 + b^2c - c^3]$.
$= \frac{3}{2bc} [a^2(b+c) + bc(b+c) - (b+c)(b^2-bc+c^2)]$.
$= \frac{3(b+c)}{2bc} [a^2 + bc - b^2 + bc - c^2] = \frac{3(b+c)}{2bc} [a^2 - (b-c)^2]$.
Substituting the values: $\frac{3(6+9)}{2(6)(9)} [57 - (6-9)^2] = \frac{3(15)}{108} [57 - 9] = \frac{45}{108} \times 48 = \frac{45 \times 4}{9} = 5 \times 4 = 20$.

(A) For Reason $(R)$: Given $r:R:r_2 = 1:2.5:6 = 2:5:12$.
Let $r=2k, R=5k, r_2=12k$.
Using the formula $r_2-r = 4R \sin^2 \frac{B}{2}$,we have $12k-2k = 4(5k) \sin^2 \frac{B}{2}$.
$10k = 20k \sin^2 \frac{B}{2}$ $\Rightarrow \sin^2 \frac{B}{2} = \frac{1}{2}$ $\Rightarrow \frac{B}{2} = 45^{\circ}$ $\Rightarrow B = 90^{\circ}$.
Thus,Reason $(R)$ is true.
For Assertion $(A)$: Given $r=6, r_2=36, R=15$.
Using $r_2-r = 4R \sin^2 \frac{B}{2}$,we have $36-6 = 4(15) \sin^2 \frac{B}{2}$.
$30 = 60 \sin^2 \frac{B}{2}$ $\Rightarrow \sin^2 \frac{B}{2} = \frac{1}{2}$ $\Rightarrow B = 90^{\circ}$.
If $B=90^{\circ}$,then $b^2 = a^2+c^2$. Thus,Assertion $(A)$ is true.
Since $(R)$ provides the logical basis for the property used in $(A)$,$(R)$ is the correct explanation of $(A)$.

(C) Given $a, b, c$ are in $AP$,so $2b = a + c$. By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$2 \sin B = \sin A + \sin C$.
Given $A = 2C$,so $2 \sin B = \sin 2C + \sin C$.
Since $A + B + C = 180^{\circ}$,we have $B = 180^{\circ} - (A + C) = 180^{\circ} - 3C$.
So,$2 \sin(180^{\circ} - 3C) = \sin 2C + \sin C$.
$2 \sin 3C = 2 \sin C \cos C + \sin C$.
$2(3 \sin C - 4 \sin^3 C) = \sin C(2 \cos C + 1)$.
Since $\sin C \neq 0$,we have $2(3 - 4 \sin^2 C) = 2 \cos C + 1$.
$6 - 8(1 - \cos^2 C) = 2 \cos C + 1$.
$6 - 8 + 8 \cos^2 C = 2 \cos C + 1$.
$8 \cos^2 C - 2 \cos C - 3 = 0$.
$(4 \cos C + 3)(2 \cos C - 1) = 0$.
Since $A = 2C$,$C < 90^{\circ}$,so $\cos C > 0$. Thus,$\cos C = \frac{1}{2}$,which means $C = 60^{\circ}$.
Then $A = 120^{\circ}$,which is impossible as $A+C < 180^{\circ}$.
Wait,re-evaluating: $8 \cos^2 C - 2 \cos C - 3 = 0$ gives $\cos C = \frac{2 \pm \sqrt{4 + 96}}{16} = \frac{2 \pm 10}{16}$.
Taking $\cos C = \frac{12}{16} = \frac{3}{4}$.
Then $\sin C = \sqrt{1 - (\frac{3}{4})^2} = \frac{\sqrt{7}}{4}$.
$B = 180^{\circ} - 3C$,so $\sin B = \sin 3C = 3 \sin C - 4 \sin^3 C = \sin C(3 - 4 \sin^2 C) = \sin C(3 - 4(1 - \cos^2 C)) = \sin C(4 \cos^2 C - 1)$.
$\sin B = \frac{\sqrt{7}}{4} (4(\frac{9}{16}) - 1) = \frac{\sqrt{7}}{4} (\frac{9}{4} - 1) = \frac{\sqrt{7}}{4} \times \frac{5}{4} = \frac{5\sqrt{7}}{16}$.
Finally,$b:c = \sin B : \sin C = \frac{5\sqrt{7}}{16} : \frac{\sqrt{7}}{4} = 5:4$.

(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.
By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.
$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\frac{\pi}{6})$.
$a^2 = 49 + 48 - 56\sqrt{3} \times \frac{\sqrt{3}}{2} = 97 - 84 = 13$.
So,$a = \sqrt{13}$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Thus,$\sin B = \frac{b \sin A}{a} = \frac{7 \sin(\pi/6)}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.
And $\sin C = \frac{c \sin A}{a} = \frac{4\sqrt{3} \sin(\pi/6)}{\sqrt{13}} = \frac{2\sqrt{3}}{\sqrt{13}}$.
Therefore,$a \sin B \sin C = \sqrt{13} \times \frac{7}{2\sqrt{13}} \times \frac{2\sqrt{3}}{\sqrt{13}} = \frac{7\sqrt{3}}{\sqrt{13}}$.

(B) $L.H.S. = (b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right)$
$= (b^2 + c^2 + 2bc) \sin^2\left(\frac{A}{2}\right) + (b^2 + c^2 - 2bc) \cos^2\left(\frac{A}{2}\right)$
$= (b^2 + c^2) \left[\sin^2\left(\frac{A}{2}\right) + \cos^2\left(\frac{A}{2}\right)\right] - 2bc \left[\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)\right]$
$= b^2 + c^2 - 2bc \cos A$
$= a^2$ (Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$)
Since $a = 2R \sin A$,we have $a^2 = 4R^2 \sin^2 A$
$= 4R^2 \left(\frac{1 - \cos 2A}{2}\right)$
$= 2R^2 (1 - \cos 2A)$
Comparing with $K(1 - \cos 2A)$,we get $K = 2R^2$.

(A) Given $A=\frac{\pi}{3}$ and $B=\frac{\pi}{4}$. In $\triangle ABC$,$\angle C = \pi - (A+B) = \pi - (\frac{\pi}{3} + \frac{\pi}{4}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin(\frac{\pi}{3}) = k \frac{\sqrt{3}}{2}$,$b = k \sin(\frac{\pi}{4}) = k \frac{1}{\sqrt{2}}$,and $c = k \sin(\frac{5\pi}{12}) = k \sin(\frac{\pi}{4} + \frac{\pi}{6}) = k (\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Now,$\frac{a^2-b^2}{c^2} = \frac{k^2(\frac{3}{4} - \frac{1}{2})}{k^2(\frac{(\sqrt{3}+1)^2}{8})} = \frac{1/4}{(\frac{3+1+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{4+2\sqrt{3}}{8})} = \frac{1/4}{(\frac{2+\sqrt{3}}{4})} = \frac{1}{2+\sqrt{3}}$.
Rationalizing the denominator: $\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = 2-\sqrt{3}$.
Thus,the correct option is $A$.

(B) Given $\frac{a}{\tan A} = \frac{b}{\tan B} = \frac{c}{\tan C}$.
Since $\tan A = \frac{\sin A}{\cos A}$,we have $\frac{a \cos A}{\sin A} = \frac{b \cos B}{\sin B} = \frac{c \cos C}{\sin C}$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Substituting this into the given equation,we get $2R \cos A = 2R \cos B = 2R \cos C$.
This implies $\cos A = \cos B = \cos C$.
Since $A, B, C$ are angles of a triangle,$A = B = C = 60^{\circ}$.
Therefore,$\cos^2 60^{\circ} + \cos^2 60^{\circ} + \cos^2 60^{\circ} = (\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

(D) Given,$\angle B = \frac{\pi}{4}$,$\angle C = \frac{\pi}{3}$.
In $\triangle ABC$,$\angle A = \pi - (\frac{\pi}{4} + \frac{\pi}{3}) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} = 75^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$.
The area of the triangle is $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} (2R \sin B)(2R \sin C) \sin A = 2R^2 \sin A \sin B \sin C$.
Given $\Delta = 54 + 18\sqrt{3} = 18(3 + \sqrt{3})$.
$\sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
$\Delta = 2R^2 \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right) = 2R^2 \left(\frac{\sqrt{3} + 1}{4\sqrt{2}}\right) \left(\frac{\sqrt{3}}{\sqrt{2}}\right) = 2R^2 \frac{3 + \sqrt{3}}{8} = R^2 \frac{3 + \sqrt{3}}{4}$.
Equating the areas: $R^2 \frac{3 + \sqrt{3}}{4} = 18(3 + \sqrt{3})$ $\Rightarrow R^2 = 72$ $\Rightarrow R = 6\sqrt{2}$.
Finally,$a = 2R \sin A = 2(6\sqrt{2}) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = 12\sqrt{2} \left(\frac{\sqrt{2}(\sqrt{3} + 1)}{4}\right) = 3(2)(\sqrt{3} + 1) = 6(\sqrt{3} + 1)$.

(A) We have to find the value of $\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2}$.
Using the sine rule,$b = 2R \sin B$ and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the expression:
$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B + \sin C))^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2R(\sin B - \sin C))^2}$
$= \frac{1}{4R^2} \left[ \frac{\cos^2 \left( \frac{B - C}{2} \right)}{(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2})^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(2 \cos \frac{B+C}{2} \sin \frac{B-C}{2})^2} \right]$
$= \frac{1}{4R^2} \left[ \frac{1}{4 \sin^2 \frac{B+C}{2}} + \frac{1}{4 \cos^2 \frac{B+C}{2}} \right]$
$= \frac{1}{16R^2} \left[ \frac{\cos^2 \frac{B+C}{2} + \sin^2 \frac{B+C}{2}}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right]$
$= \frac{1}{16R^2} \left[ \frac{1}{\sin^2 \frac{B+C}{2} \cos^2 \frac{B+C}{2}} \right] = \frac{1}{4R^2} \left[ \frac{1}{\sin^2 (B+C)} \right]$
Since $A+B+C = \pi$,$\sin(B+C) = \sin(\pi - A) = \sin A$.
Thus,the expression becomes $\frac{1}{4R^2 \sin^2 A} = \frac{1}{(2R \sin A)^2} = \frac{1}{a^2}$.

(D) Given the expression $\frac{1+\cos(A-B) \cdot \cos C}{1+\cos(A-C) \cdot \cos B}$.
Since $A+B+C = 180^{\circ}$,we have $C = 180^{\circ} - (A+B)$ and $B = 180^{\circ} - (A+C)$.
Thus,$\cos C = -\cos(A+B)$ and $\cos B = -\cos(A+C)$.
Substituting these into the expression:
$= \frac{1 - \cos(A-B)\cos(A+B)}{1 - \cos(A-C)\cos(A+C)}$
Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$:
$= \frac{1 - (\cos^2 A - \sin^2 B)}{1 - (\cos^2 A - \sin^2 C)}$
$= \frac{1 - \cos^2 A + \sin^2 B}{1 - \cos^2 A + \sin^2 C}$
$= \frac{\sin^2 A + \sin^2 B}{\sin^2 A + \sin^2 C}$
Using the Sine Rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$.
$= \frac{(a/k)^2 + (b/k)^2}{(a/k)^2 + (c/k)^2} = \frac{a^2+b^2}{a^2+c^2}$.

(A) Given that in $\triangle ABC$,$\angle C = 90^{\circ}$.
Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle A + \angle B = 90^{\circ}$.
Using the sine rule,$a = k \sin A$ and $b = k \sin B$.
Substituting these into the expression:
$\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = \frac{k^2 \sin^2 A + k^2 \sin^2 B}{k^2 \sin^2 A - k^2 \sin^2 B} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin(A-B)$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B) = \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$.
Since $A+B = 90^{\circ}$,$\sin(A+B) = 1$ and $B = 90^{\circ}-A$,so $\sin B = \cos A$.
$= \frac{\sin^2 A + \cos^2 A}{1} = 1$.

(D) Let $\angle B = \theta$. Then $\angle A = 3\theta$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle C = 180^{\circ} - (A + B) = 180^{\circ} - 4\theta$.
Using the sine rule,$\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}$.
Substituting the given values: $\frac{AB}{\sin(180^{\circ} - 4\theta)} = \frac{16}{\sin 3\theta} = \frac{9}{\sin \theta}$.
From $\frac{16}{\sin 3\theta} = \frac{9}{\sin \theta}$,we have $16 \sin \theta = 9 \sin 3\theta = 9(3 \sin \theta - 4 \sin^3 \theta)$.
Dividing by $\sin \theta$ (since $\sin \theta \neq 0$),$16 = 27 - 36 \sin^2 \theta$,which gives $36 \sin^2 \theta = 11$,so $\sin^2 \theta = \frac{11}{36}$.
Then $\cos^2 \theta = 1 - \frac{11}{36} = \frac{25}{36}$,so $\cos \theta = \frac{5}{6}$.
Now,$\sin 4\theta = 2 \sin 2\theta \cos 2\theta = 4 \sin \theta \cos \theta (1 - 2 \sin^2 \theta) = 4 \times \frac{\sqrt{11}}{6} \times \frac{5}{6} \times (1 - 2 \times \frac{11}{36}) = \frac{20\sqrt{11}}{36} \times (1 - \frac{11}{18}) = \frac{5\sqrt{11}}{9} \times \frac{7}{18} = \frac{35\sqrt{11}}{162}$.
Finally,$AB = \frac{9 \sin 4\theta}{\sin \theta} = 9 \times \frac{35\sqrt{11}}{162} \times \frac{6}{\sqrt{11}} = \frac{35}{3}$.

(C) Let a triangle $ABC$ have sides $a, b, c$ and area $\Delta$.
We know that the area $\Delta = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C$.
From the cosine rule,we have $a^2 = b^2+c^2-2bc \cos A$,$b^2 = a^2+c^2-2ac \cos B$,and $c^2 = a^2+b^2-2ab \cos C$.
Adding these three equations gives $a^2+b^2+c^2 = 2(a^2+b^2+c^2) - 2(bc \cos A + ac \cos B + ab \cos C)$.
Rearranging,we get $a^2+b^2+c^2 = 2(bc \cos A + ac \cos B + ab \cos C)$.
Since $\Delta = \frac{1}{2}bc \sin A$,we have $bc = \frac{2\Delta}{\sin A}$. Similarly,$ac = \frac{2\Delta}{\sin B}$ and $ab = \frac{2\Delta}{\sin C}$.
Substituting these into the equation:
$a^2+b^2+c^2 = 2 \left( \frac{2\Delta}{\sin A} \cos A + \frac{2\Delta}{\sin B} \cos B + \frac{2\Delta}{\sin C} \cos C \right)$
$a^2+b^2+c^2 = 4\Delta (\cot A + \cot B + \cot C)$
Therefore,$\cot A + \cot B + \cot C = \frac{a^2+b^2+c^2}{4\Delta}$.

(C) Using the Napier's analogy,we have $\tan \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}$ and $\tan \frac{A+B}{2} = \cot \frac{C}{2}$.
Given $\tan \frac{A-B}{2} = \frac{1}{4} \tan \frac{A+B}{2}$,we substitute these values:
$\frac{a-b}{a+b} \cot \frac{C}{2} = \frac{1}{4} \cot \frac{C}{2}$
$\Rightarrow \frac{a-b}{a+b} = \frac{1}{4}$
$\Rightarrow 4(a-b) = a+b$
$\Rightarrow 4a - 4b = a + b$
$\Rightarrow 3a = 5b$
Given $a=5$,we have $3(5) = 5b \Rightarrow b=3$.
Now,$\sqrt{a^2-b^2} = \sqrt{5^2-3^2} = \sqrt{25-9} = \sqrt{16} = 4$.

(A) In $\triangle ABC$,let $BD$ be the median drawn from vertex $B$ to side $AC$.
Given sides are $a = BC = 5$,$b = AC = 6$,and $c = AB = 7$.
The formula for the length of the median $m_b$ drawn from vertex $B$ is given by:
$BD = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}$
Substituting the given values:
$BD = \frac{1}{2} \sqrt{2(5)^2 + 2(7)^2 - (6)^2}$
$BD = \frac{1}{2} \sqrt{2(25) + 2(49) - 36}$
$BD = \frac{1}{2} \sqrt{50 + 98 - 36}$
$BD = \frac{1}{2} \sqrt{112}$
$BD = \frac{1}{2} \sqrt{16 \times 7}$
$BD = \frac{1}{2} \times 4 \sqrt{7}$
$BD = 2 \sqrt{7}$

(B) Let the angles of the triangle be $x, 2x$,and $3x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
Thus,the angles are $30^{\circ}, 60^{\circ}$,and $90^{\circ}$.
According to the Sine Rule,the sides $a, b, c$ are proportional to the sines of their opposite angles: $a: b: c = \sin A: \sin B: \sin C$.
Substituting the angles: $a: b: c = \sin 30^{\circ}: \sin 60^{\circ}: \sin 90^{\circ}$.
$a: b: c = \frac{1}{2}: \frac{\sqrt{3}}{2}: 1$.
Multiplying by $2$,we get the ratio $1: \sqrt{3}: 2$.

(A) Let the sides of the triangle be $a = k$,$b = \sqrt{3}k$,and $c = 2k$.
Since $a^2 + b^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2 = c^2$,the triangle is a right-angled triangle with the hypotenuse $c = 2k$.
Let the angles opposite to sides $a, b, c$ be $A, B, C$ respectively.
Then $C = 90^{\circ}$.
Using the sine rule or trigonometric ratios in a right triangle:
$\sin A = \frac{a}{c} = \frac{k}{2k} = \frac{1}{2} \implies A = 30^{\circ}$.
$\sin B = \frac{b}{c} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \implies B = 60^{\circ}$.
Thus,the angles are $30^{\circ}, 60^{\circ}, 90^{\circ}$.
The ratio of the angles is $30^{\circ} : 60^{\circ} : 90^{\circ} = 1 : 2 : 3$.

(A) Given,$8R^2 = a^2 + b^2 + c^2$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these in the given equation:
$8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$
$8R^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C)$
$2 = \sin^2 A + \sin^2 B + \sin^2 C$
Using $\sin^2 \theta = 1 - \cos^2 \theta$:
$2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$
$2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$
$\cos^2 A + \cos^2 B = \sin^2 C$
$\cos^2 A + \cos^2 B = 1 - \cos^2 C$
$\cos^2 A + \cos^2 B + \cos^2 C = 1$
This is a known identity for a right-angled triangle where one of the angles is $90^\circ$.
Alternatively,$\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C = 1$ implies $\cos A \cos B \cos C = 0$.
Thus,$A = 90^\circ$ or $B = 90^\circ$ or $C = 90^\circ$.
Therefore,the triangle is a right-angled triangle.

(C) Given,$\angle A=90^{\circ}$ and $\angle B \neq \angle C$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $b = k \sin B$ and $c = k \sin C$.
Substituting these into the expression:
$\frac{b^2+c^2}{b^2-c^2} \sin (B-C) = \frac{k^2 \sin^2 B + k^2 \sin^2 C}{k^2 \sin^2 B - k^2 \sin^2 C} \sin (B-C)$
$= \frac{\sin^2 B + \sin^2 C}{\sin^2 B - \sin^2 C} \sin (B-C)$
Since $\angle A = 90^{\circ}$,then $B+C = 90^{\circ}$,so $C = 90^{\circ}-B$.
Thus,$\sin C = \cos B$ and $\sin^2 C = \cos^2 B$.
Also,$\sin^2 B - \sin^2 C = \sin(B+C) \sin(B-C) = \sin(90^{\circ}) \sin(B-C) = 1 \cdot \sin(B-C)$.
Substituting these:
$= \frac{\sin^2 B + \cos^2 B}{\sin(B+C)} \cdot \sin(B-C) = \frac{1}{\sin(90^{\circ})} = 1$.

(D) Given,the ratio of angles of a triangle is $1: 1: 4$. Let the angles be $A, B$,and $C$.
$\therefore A: B: C = 1: 1: 4$
Let $A = x, B = x$,and $C = 4x$.
Since $A + B + C = 180^{\circ}$,we have $x + x + 4x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
Thus,$A = 30^{\circ}, B = 30^{\circ}$,and $C = 120^{\circ}$.
The largest angle is $120^{\circ}$,so the largest side is $c$.
The ratio of the perimeter to the largest side is $(a + b + c) : c$.
Using the sine rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.
Ratio $= (2R \sin 30^{\circ} + 2R \sin 30^{\circ} + 2R \sin 120^{\circ}) : 2R \sin 120^{\circ}$
$= (\sin 30^{\circ} + \sin 30^{\circ} + \sin 120^{\circ}) : \sin 120^{\circ}$
$= (\frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2}$
$= (1 + \frac{\sqrt{3}}{2}) : \frac{\sqrt{3}}{2} = (2 + \sqrt{3}) : \sqrt{3}$.

(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k$ is a constant.
Substituting these values into the given equation:
$\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$
$\Rightarrow \cot A = \cot B = \cot C$
Since $A, B, C$ are angles of a triangle,$A = B = C$.
Since all angles are equal,the triangle is an equilateral triangle.

(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.
Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} = k$.
Then,$\frac{a^2-b^2}{a^2+b^2} = \frac{k^2 \sin^2 A - k^2 \sin^2 B}{k^2 \sin^2 A + k^2 \sin^2 B} = \frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B}$.
Since $B = 90^{\circ} - A$,we have $\sin B = \cos A$ and $\cos B = \sin A$.
Substituting these,$\frac{\sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A} = \frac{-(\cos^2 A - \sin^2 A)}{1} = -\cos 2A$.
Alternatively,using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:
$\frac{\sin(A+B)\sin(A-B)}{\sin^2 A + \cos^2 A} = \sin(90^{\circ})\sin(A-B) = 1 \cdot \sin(A-B) = \sin(A-B)$.

(B) Given: $b=20, c=21$ and $\sin A=\frac{3}{5}$.
Using the identity $\cos^2 A = 1 - \sin^2 A$,we get:
$\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Thus,$\cos A = \frac{4}{5}$ (assuming $A$ is an acute angle).
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting the values: $\frac{4}{5} = \frac{20^2+21^2-a^2}{2 \times 20 \times 21}$.
$\frac{4}{5} = \frac{400+441-a^2}{840}$.
$840 \times \frac{4}{5} = 841 - a^2$.
$168 \times 4 = 841 - a^2$.
$672 = 841 - a^2$.
$a^2 = 841 - 672 = 169$.
Therefore,$a = \sqrt{169} = 13$.

(C) Using the Law of Cosines,we find the values of $\cos A, \cos B, \cos C$:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25+49-9}{2(5)(7)} = \frac{65}{70} = \frac{13}{14}$
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{9+49-25}{2(3)(7)} = \frac{33}{42} = \frac{11}{14}$
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9+25-49}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2}$
Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{3+5+7}{2} = 7.5 = \frac{15}{2}$.
$\Delta = \sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)} = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} = \sqrt{\frac{675}{16}} = \frac{15\sqrt{3}}{4}$.
Since $\cot A = \frac{\cos A}{\sin A} = \frac{\cos A}{\Delta / (\frac{1}{2}bc)} = \frac{2bc \cos A}{4\Delta} = \frac{b^2+c^2-a^2}{4\Delta}$,we have:
$\cot A + \cot B + \cot C = \frac{b^2+c^2-a^2 + a^2+c^2-b^2 + a^2+b^2-c^2}{4\Delta} = \frac{a^2+b^2+c^2}{4\Delta}$
$= \frac{9+25+49}{4(\frac{15\sqrt{3}}{4})} = \frac{83}{15\sqrt{3}}$.

(D) Given $a=4, b=3, c=2$ in $\triangle ABC$.
Using the Law of Cosines:
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{16+9-4}{2 \times 4 \times 3} = \frac{21}{24} = \frac{7}{8}$.
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{16+4-9}{2 \times 4 \times 2} = \frac{11}{16} \implies \sec B = \frac{16}{11}$.
Now,substitute these values into the expression:
$2(a-b \cos C)(a-c \sec B) = 2(4 - 3 \times \frac{7}{8})(4 - 2 \times \frac{16}{11})$
$= 2(4 - \frac{21}{8})(4 - \frac{32}{11})$
$= 2(\frac{32-21}{8})(\frac{44-32}{11})$
$= 2(\frac{11}{8})(\frac{12}{11}) = 2 \times \frac{12}{8} = 2 \times \frac{3}{2} = 3$.

(B) We know that $r_1 = \frac{\Delta}{s-a} = 6 \Rightarrow s-a = \frac{\Delta}{6}$ $(i)$
$r_2 = \frac{\Delta}{s-b} = 9 \Rightarrow s-b = \frac{\Delta}{9}$ $(ii)$
$r_3 = \frac{\Delta}{s-c} = 18 \Rightarrow s-c = \frac{\Delta}{18}$ $(iii)$
Adding $(i), (ii)$ and $(iii)$ gives $3s - (a+b+c) = \Delta(\frac{1}{6} + \frac{1}{9} + \frac{1}{18}) = \Delta(\frac{3+2+1}{18}) = \frac{\Delta}{3}$.
Since $a+b+c = 2s$,we have $3s - 2s = \frac{\Delta}{3} \Rightarrow s = \frac{\Delta}{3}$.
Substituting $s$ in $(i), (ii), (iii)$:
$a = s - \frac{\Delta}{6} = \frac{\Delta}{3} - \frac{\Delta}{6} = \frac{\Delta}{6} = \frac{3\Delta}{18}$.
$b = s - \frac{\Delta}{9} = \frac{\Delta}{3} - \frac{\Delta}{9} = \frac{2\Delta}{9} = \frac{4\Delta}{18}$.
$c = s - \frac{\Delta}{18} = \frac{\Delta}{3} - \frac{\Delta}{18} = \frac{5\Delta}{18}$.
Thus,$a:b:c = 3:4:5$. Let $a=3k, b=4k, c=5k$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(4k)^2+(5k)^2-(3k)^2}{2(4k)(5k)} = \frac{(16+25-9)k^2}{40k^2} = \frac{32}{40} = \frac{4}{5}$.

(B) Given,in $\triangle ABC$,$A$ is acute,$C$ is obtuse,$\sin A = \frac{3\sqrt{3}}{14}$,$a = 3$,and $b = 5$.
First,calculate $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3\sqrt{3}}{14}\right)^2} = \sqrt{1 - \frac{27}{196}} = \sqrt{\frac{169}{196}} = \frac{13}{14}$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\frac{13}{14} = \frac{5^2 + c^2 - 3^2}{2 \times 5 \times c} = \frac{25 + c^2 - 9}{10c} = \frac{16 + c^2}{10c}$.
Cross-multiplying: $130c = 14(16 + c^2) \Rightarrow 130c = 224 + 14c^2$.
Rearranging: $14c^2 - 130c + 224 = 0 \Rightarrow 7c^2 - 65c + 112 = 0$.
Factoring: $7c^2 - 49c - 16c + 112 = 0$ $\Rightarrow 7c(c - 7) - 16(c - 7) = 0$ $\Rightarrow (7c - 16)(c - 7) = 0$.
So,$c = \frac{16}{7}$ or $c = 7$.
Since $C$ is obtuse,the side $c$ must be the longest side.
Checking the conditions: If $c = \frac{16}{7} \approx 2.28$,then $a=3, b=5, c=2.28$. Here $b$ is the longest side,so $B$ would be obtuse.
If $c = 7$,then $a=3, b=5, c=7$. Here $c$ is the longest side,so $C$ is obtuse.
Therefore,$c = 7$.