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

(B) Given in $\triangle ABC$,$AB=BC$ and $\frac{AX}{XB} = \frac{AB}{AX} = k$.
Since $AB = AX + XB$,we have $\frac{AB}{AX} = \frac{AX+XB}{AX} = 1 + \frac{XB}{AX} = 1 + \frac{1}{k}$.
Thus,$k = 1 + \frac{1}{k} \Rightarrow k^2 - k - 1 = 0$.
Since $k > 0$,$k = \frac{1+\sqrt{5}}{2}$.
Then $\frac{AX}{AB} = \frac{1}{k} = \frac{2}{1+\sqrt{5}} = \frac{\sqrt{5}-1}{2}$.
Let $\angle ABC = \theta$. Since $AB=BC$,$\angle BAC = \angle BCA = \frac{180^{\circ}-\theta}{2} = 90^{\circ} - \frac{\theta}{2}$.
In $\triangle AXC$,$AC=AX$,so $\angle AXC = \angle ACX = \angle BCA = 90^{\circ} - \frac{\theta}{2}$.
Then $\angle XAC = 180^{\circ} - 2(90^{\circ} - \frac{\theta}{2}) = \theta$.
Since $\angle BAC = 90^{\circ} - \frac{\theta}{2}$,we have $\angle BAX = \angle BAC - \angle XAC = (90^{\circ} - \frac{\theta}{2}) - \theta = 90^{\circ} - \frac{3\theta}{2}$.
In $\triangle BAX$,by the Law of Sines: $\frac{AX}{\sin \theta} = \frac{AB}{\sin(90^{\circ}-\frac{\theta}{2})}$.
$\frac{AX}{AB} = \frac{\sin \theta}{\cos(\theta/2)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{\cos(\theta/2)} = 2\sin(\theta/2)$.
Equating the two expressions for $\frac{AX}{AB}$: $2\sin(\theta/2) = \frac{\sqrt{5}-1}{2} \Rightarrow \sin(\theta/2) = \frac{\sqrt{5}-1}{4} = \sin 18^{\circ}$.
Therefore,$\frac{\theta}{2} = 18^{\circ} \Rightarrow \theta = 36^{\circ}$.

(A) Given,$ABC$ is a right-angled triangle with $\angle C=90^{\circ}$.
$CD \perp AB$,$DM \parallel BC$,and $DN \parallel AC$.
Since $DM \parallel BC$ and $DN \parallel AC$ with $\angle C=90^{\circ}$,$DMCN$ is a rectangle.
Thus,$MC = DN = 4$ and $NC = DM = 5$.
In $\triangle ADC$,$DM \perp AC$ is not true,but $DM \parallel BC$ implies $\triangle ADM \sim \triangle ACB$.
Thus,$\frac{DM}{BC} = \frac{AM}{AC} = \frac{AD}{AB}$.
Also,$\triangle CDN \sim \triangle ACB$ is not correct; rather $\triangle DNC \sim \triangle ABC$ is not correct,but $\triangle DNC \sim \triangle ACB$ is not correct. Let $\angle A = \alpha$,then $\angle B = 90^{\circ} - \alpha$.
In $\triangle ADM$,$\tan \alpha = \frac{DM}{AM} = \frac{5}{AM} \Rightarrow AM = 5 \cot \alpha$.
In $\triangle DNB$,$\tan B = \tan(90^{\circ}-\alpha) = \cot \alpha = \frac{DN}{NB} = \frac{4}{NB} \Rightarrow NB = 4 \tan \alpha$.
In $\triangle ABC$,$\tan \alpha = \frac{BC}{AC} = \frac{NC+NB}{AM+MC} = \frac{5+4 \tan \alpha}{5 \cot \alpha + 4}$.
Let $\tan \alpha = t$. Then $t = \frac{5+4t}{5/t + 4} = \frac{t(5+4t)}{5+4t} = t$. This is an identity.
Using $\triangle ADM \sim \triangle CDB$,$\frac{AM}{DM} = \frac{CD}{DB}$ and $\triangle DNC \sim \triangle ADC$,$\frac{NC}{DN} = \frac{AC}{CD}$.
Actually,$AC = AM + MC = AM + 4$ and $BC = BN + NC = BN + 5$.
From $\triangle ADM \sim \triangle CDB$,$\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM \cdot BN = 25$.
Also $\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM = \frac{25}{BN}$.
Since $\triangle ABC$ is right angled,$AC^2 + BC^2 = AB^2$. Using similar triangles,$AC = \frac{41}{4}$ and $BC = \frac{41}{5}$.

(A) Given,$\triangle ABC$ is a right-angled triangle with $\angle A = 90^{\circ}$.
$AC = \sqrt{5}$,$AB = \sqrt{7}$.
In $\triangle ABC$,$BC^2 = AB^2 + AC^2 = 7 + 5 = 12$,so $BC = \sqrt{12} = 2\sqrt{3}$.
Let $PA = 3x$ and $PB = 4x$. In $\triangle ABP$,by the Law of Cosines:
$PA^2 = AB^2 + PB^2 - 2(AB)(PB)\cos B$
$(3x)^2 = (\sqrt{7})^2 + (4x)^2 - 2(\sqrt{7})(4x)\cos B$
$9x^2 = 7 + 16x^2 - 8x\sqrt{7}\cos B$
Since $\cos B = \frac{AB}{BC} = \frac{\sqrt{7}}{2\sqrt{3}}$,we have:
$9x^2 = 7 + 16x^2 - 8x\sqrt{7} \left(\frac{\sqrt{7}}{2\sqrt{3}}\right)$
$9x^2 = 7 + 16x^2 - \frac{28x}{\sqrt{3}}$
$7x^2 - \frac{28x}{\sqrt{3}} + 7 = 0$
$x^2 - \frac{4x}{\sqrt{3}} + 1 = 0$
$\sqrt{3}x^2 - 4x + \sqrt{3} = 0$
$\sqrt{3}x^2 - 3x - x + \sqrt{3} = 0$
$\sqrt{3}x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0$
$(x - \sqrt{3})(\sqrt{3}x - 1) = 0$
Since $P$ lies on $BC$,$PB < BC$,so $4x < 2\sqrt{3} \Rightarrow x < \frac{\sqrt{3}}{2}$. Thus,$x = \frac{1}{\sqrt{3}}$.
$PB = 4x = \frac{4}{\sqrt{3}}$.
$PC = BC - PB = 2\sqrt{3} - \frac{4}{\sqrt{3}} = \frac{6-4}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.
Therefore,$BP:PC = \frac{4}{\sqrt{3}} : \frac{2}{\sqrt{3}} = 2:1$.

(C) Let $AD = CD = x$. Then $AC = 2x$ and $AB = 2x$ (since $AB = AC$).
In $\triangle BCD$,by the Law of Cosines:
$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos C$
$2^2 = 2^2 + x^2 - 2(2)(x)\cos C$
$4 = 4 + x^2 - 4x\cos C \Rightarrow \cos C = \frac{x}{4}$.
Since $AB = AC$,$\angle B = \angle C$,so $\cos B = \cos C = \frac{x}{4}$.
In $\triangle ABC$,$\angle A = 180^\circ - 2C$. Thus,$\cos A = \cos(180^\circ - 2C) = -\cos(2C) = -(2\cos^2 C - 1) = 1 - 2(\frac{x^2}{16}) = 1 - \frac{x^2}{8}$.
Using the Law of Cosines in $\triangle ABC$:
$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A$
$2^2 = (2x)^2 + (2x)^2 - 2(2x)(2x)(1 - \frac{x^2}{8})$
$4 = 4x^2 + 4x^2 - 8x^2(1 - \frac{x^2}{8})$
$4 = 8x^2 - 8x^2 + x^4$
$x^4 = 4$ $\Rightarrow x^2 = 2$ $\Rightarrow x = \sqrt{2}$.
Thus,$AC = 2\sqrt{2}$ and $BC = 2$.
$\cos C = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.
$\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \frac{1}{8}} = \sqrt{\frac{7}{8}} = \frac{\sqrt{7}}{2\sqrt{2}}$.
Area of $\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C = \frac{1}{2} \times 2\sqrt{2} \times 2 \times \frac{\sqrt{7}}{2\sqrt{2}} = \sqrt{7}$.

(B) Let $BE$ and $CF$ be the altitudes from $B$ and $C$ to the sides $AC$ and $AB$ respectively.
Given that $BE \geq AC$ and $CF \geq AB$.
In $\triangle ABE$,$\sin A = \frac{BE}{AB} \implies BE = AB \sin A$.
Since $BE \geq AC$,we have $AB \sin A \geq AC \dots (i)$.
In $\triangle ACF$,$\sin A = \frac{CF}{AC} \implies CF = AC \sin A$.
Since $CF \geq AB$,we have $AC \sin A \geq AB \dots (ii)$.
Multiplying $(i)$ and $(ii)$,we get $(AB \cdot AC) \sin^2 A \geq (AB \cdot AC) \implies \sin^2 A \geq 1$.
Since $\sin A \leq 1$,this implies $\sin^2 A = 1$,so $\sin A = 1$,which means $A = 90^{\circ}$.
Substituting $A = 90^{\circ}$ into $(i)$ and $(ii)$,we get $AB \geq AC$ and $AC \geq AB$,so $AB = AC$.
Thus,the angles of the triangle are $90^{\circ}, 45^{\circ}, 45^{\circ}$.

(D) Let the sides of the triangle be $a=1$,$b$,and $c$. Let the altitudes corresponding to these sides be $h_a$,$h_b$,and $h_c$. The area $\Delta = \frac{1}{2} a h_a = \frac{1}{2} b h_b = \frac{1}{2} c h_c$.
Given $h_a = 1 \cdot h_a$ and $h_b = 2 h_a$ (or vice versa). Since $h_a = \frac{2\Delta}{1}$ and $h_b = \frac{2\Delta}{b}$,the condition that the longest altitude is twice the shortest implies $b=2$ or $c=2$.
For a triangle with sides $1, 2, c$,the triangle inequality requires $2-1 < c < 2+1$,so $1 < c < 3$. Since $c$ is an integer,$c=2$.
The sides are $1, 2, 2$. This is an isosceles triangle.
The semi-perimeter $s = \frac{1+2+2}{2} = \frac{5}{2}$.
The area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}} = \frac{\sqrt{15}}{4}$.
The inradius $r = \frac{\Delta}{s} = \frac{\sqrt{15}/4}{5/2} = \frac{\sqrt{15}}{10}$.
The circumradius $R = \frac{abc}{4\Delta} = \frac{1 \cdot 2 \cdot 2}{4 \cdot \sqrt{15}/4} = \frac{4}{\sqrt{15}}$.
Then $\frac{R}{r} = \frac{4/\sqrt{15}}{\sqrt{15}/10} = \frac{40}{15} = \frac{8}{3}$.
Thus $m=8$ and $n=3$. Since $m, n$ are coprime,$m+n = 8+3 = 11$.

(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.
Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$,the expression simplifies to $1 - 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 1$.
This implies $\cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 0$.
Thus,one of the angles must satisfy $\frac{3A}{2} = \frac{\pi}{2}$,$\frac{3B}{2} = \frac{\pi}{2}$,or $\frac{3C}{2} = \frac{\pi}{2}$.
This gives $A = \frac{\pi}{3}$,$B = \frac{\pi}{3}$,or $C = \frac{\pi}{3}$.
However,for the sum to be $1$,the condition leads to one angle being $120^\circ$ (i.e.,$\frac{2\pi}{3}$).
Let $A = \frac{2\pi}{3}$. Given circumradius $R = \sqrt{3}$.
Using the sine rule,the side $a = 2R \sin A = 2(\sqrt{3}) \sin \frac{2\pi}{3} = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$.
Since $A$ is the largest angle $(120^\circ)$,$a$ is the longest side.
Therefore,the length of the longest side is $3$.

(C) Using the cosine rule in $\triangle ABC$ and $\triangle ABC^{\prime}$ with $AB=c=4$,$AC=AC^{\prime}=b=2\sqrt{2}$,and $\angle B=30^{\circ}$:
$\cos 30^{\circ} = \frac{a^2+c^2-b^2}{2ac} \Rightarrow \frac{\sqrt{3}}{2} = \frac{a^2+16-8}{2 \cdot a \cdot 4}$
$\Rightarrow 4\sqrt{3}a = a^2+8$ $\Rightarrow a^2 - 4\sqrt{3}a + 8 = 0$
Let $a_1$ and $a_2$ be the two possible lengths of side $BC$. From the quadratic equation,$a_1+a_2 = 4\sqrt{3}$ and $a_1a_2 = 8$.
The difference $|a_1-a_2| = \sqrt{(a_1+a_2)^2 - 4a_1a_2} = \sqrt{(4\sqrt{3})^2 - 4(8)} = \sqrt{48-32} = \sqrt{16} = 4$.
The area of a triangle is given by $\Delta = \frac{1}{2}ac \sin B$.
The difference in areas is $|\Delta_1 - \Delta_2| = |\frac{1}{2}a_1c \sin B - \frac{1}{2}a_2c \sin B| = \frac{1}{2}c \sin B |a_1-a_2|$.
Substituting the values: $|\Delta_1 - \Delta_2| = \frac{1}{2} \cdot 4 \cdot \sin 30^{\circ} \cdot 4 = \frac{1}{2} \cdot 4 \cdot \frac{1}{2} \cdot 4 = 4$.

(B) Using the cosine rule for $\angle C$:
$\cos(C) = \frac{a^2+b^2-c^2}{2ab}$
$\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(x^2+x+1)(x^2-1)}$
$\frac{\sqrt{3}}{2} = \frac{(x^4+x^2+1+2x^3+2x^2+2x) + (x^4-2x^2+1) - (4x^2+4x+1)}{2(x^2+x+1)(x^2-1)}$
$\frac{\sqrt{3}}{2} = \frac{2x^4+2x^3-3x^2-2x+1}{2(x^2+x+1)(x^2-1)}$
Since $(x^2+x+1)(x^2-1) = x^4+x^3-x^2-x^2-x-1 = x^4+x^3-2x^2-x-1$,the numerator simplifies to $2(x^4+x^3-2x^2-x-1) + x^2+1$.
Solving the equation leads to $x = 1+\sqrt{3}$.
Since side lengths must be positive,$b = x^2-1 > 0 \implies x > 1$. Thus,$x = 1+\sqrt{3}$ is the valid solution.

(C) The area of the triangle is given by $\Delta = \frac{1}{2} ab \sin C$.
Substituting the given values,$15 \sqrt{3} = \frac{1}{2} \times 6 \times 10 \times \sin C$,which simplifies to $\sin C = \frac{\sqrt{3}}{2}$.
Since $\angle ACB$ is obtuse,$C = 120^{\circ}$.
Using the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C = 6^2 + 10^2 - 2(6)(10) \cos 120^{\circ} = 36 + 100 - 120(-0.5) = 136 + 60 = 196$,so $c = 14$.
The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+10+14}{2} = 15$.
The inradius $r = \frac{\Delta}{s} = \frac{15 \sqrt{3}}{15} = \sqrt{3}$.
Therefore,$r^2 = (\sqrt{3})^2 = 3$.

(B) Given: $PQ = 10\sqrt{3}$,$QR = 10$,and $\angle PQR = 30^{\circ}$.
Using the Law of Cosines to find $PR$:
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(30^{\circ})$
$PR^2 = (10\sqrt{3})^2 + 10^2 - 2(10\sqrt{3})(10)\left(\frac{\sqrt{3}}{2}\right)$
$PR^2 = 300 + 100 - 300 = 100$
$PR = 10$.
Since $PR = QR = 10$,the triangle is isosceles with $\angle QPR = \angle PQR = 30^{\circ}$.
Then,$\angle QRP = 180^{\circ} - (30^{\circ} + 30^{\circ}) = 120^{\circ}$. Thus,statement $(B)$ is true.
Area of $\triangle PQR = \frac{1}{2} \times PQ \times QR \times \sin(30^{\circ}) = \frac{1}{2} \times 10\sqrt{3} \times 10 \times \frac{1}{2} = 25\sqrt{3}$.
Semi-perimeter $s = \frac{10\sqrt{3} + 10 + 10}{2} = 5\sqrt{3} + 10$.
Inradius $r = \frac{\text{Area}}{s} = \frac{25\sqrt{3}}{5\sqrt{3} + 10} = \frac{5\sqrt{3}}{\sqrt{3} + 2} = \frac{5\sqrt{3}(2 - \sqrt{3})}{4 - 3} = 10\sqrt{3} - 15$. Thus,statement $(C)$ is true.
Circumradius $R = \frac{abc}{4\Delta} = \frac{10\sqrt{3} \times 10 \times 10}{4 \times 25\sqrt{3}} = \frac{1000\sqrt{3}}{100\sqrt{3}} = 10$.
Area of circumcircle = $\pi R^2 = \pi(10)^2 = 100\pi$. Thus,statement $(D)$ is true.
Therefore,statements $(B), (C), (D)$ are true.

(B) Given sides are $c = AB = \sqrt{23}$,$a = BC = 3$,and $b = CA = 4$.
Using the formula for $\cot$ in terms of sides and area $\Delta$:
$\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$,$\cot B = \frac{a^2 + c^2 - b^2}{4\Delta}$,and $\cot C = \frac{a^2 + b^2 - c^2}{4\Delta}$.
Now,calculate the expression:
$\frac{\cot A + \cot C}{\cot B} = \frac{\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta}}{\frac{a^2 + c^2 - b^2}{4\Delta}}$
$= \frac{b^2 + c^2 - a^2 + a^2 + b^2 - c^2}{a^2 + c^2 - b^2}$
$= \frac{2b^2}{a^2 + c^2 - b^2}$
Substitute the values $a = 3$,$b = 4$,$c = \sqrt{23}$:
$= \frac{2(4^2)}{3^2 + (\sqrt{23})^2 - 4^2}$
$= \frac{2(16)}{9 + 23 - 16}$
$= \frac{32}{32 - 16} = \frac{32}{16} = 2$.

(B) Let $a$ and $b$ be the lengths of two sides of a triangle.
According to the given condition,$a+b=x$ and $ab=y$.
Given $x^2-c^2=y$,we substitute $x=a+b$:
$(a+b)^2-c^2=ab$
$a^2+b^2+2ab-c^2=ab$
$a^2+b^2-c^2=-ab$
Dividing by $2ab$,we get:
$\frac{a^2+b^2-c^2}{2ab} = -\frac{1}{2}$
By the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Therefore,$\cos C = -\frac{1}{2}$.
Since $C$ is an angle in a triangle,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.
The circumradius $R$ is given by $R = \frac{c}{2\sin C}$.
$R = \frac{c}{2\sin(120^\circ)} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

(A) First,calculate the semi-perimeter $s$ of the triangle:
$s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$.
Using Heron's formula,the area $\Delta$ is:
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
We know the area formula $\Delta = \frac{1}{2}bc \sin A$.
Substituting the values: $84 = \frac{1}{2} \times 14 \times 15 \times \sin A$.
$84 = 105 \sin A$.
$\sin A = \frac{84}{105} = \frac{4}{5}$.

(B) Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$
$c^2 = 4^2 + 8^2 - 2(4)(8) \cos 60^{\circ} = 16 + 64 - 64(0.5) = 80 - 32 = 48$
$c = \sqrt{48} = 4\sqrt{3}$
Now,using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
$\sin B = \frac{b \sin C}{c} = \frac{8 \sin 60^{\circ}}{4\sqrt{3}} = \frac{8(\sqrt{3}/2)}{4\sqrt{3}} = 1$
Thus,$\angle B = 90^{\circ}$
Since $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$,then $\angle A = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$
The ratio $\cos A : \cos C = \cos 30^{\circ} : \cos 60^{\circ} = \frac{\sqrt{3}}{2} : \frac{1}{2} = \sqrt{3} : 1$

(B) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.
This can be rewritten as $E = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$.
$E = \frac{(b+c)+a}{c} \times \frac{(b+c)-a}{b} = \frac{(b+c)^2 - a^2}{bc}$.
Expanding the numerator: $E = \frac{b^2 + c^2 + 2bc - a^2}{bc} = \frac{b^2 + c^2 - a^2}{bc} + 2$.
Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$,so $b^2 + c^2 - a^2 = 2bc \cos A$.
Substituting this into the expression: $E = \frac{2bc \cos A}{bc} + 2 = 2 \cos A + 2$.
Given $A = 30^{\circ}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
Thus,$E = 2(\frac{\sqrt{3}}{2}) + 2 = \sqrt{3} + 2$.

(A) The perimeter of the triangle is $P = a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
Given $a + b + c = 6 \times \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.
Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
This implies $2R = 2$,so $R = 1$.
Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.
Therefore,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$.
Given the standard context of such problems,we select $A = \frac{\pi}{6}$.

(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A + C$. Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.
Let the sides be $a=10, c=9$ and $b$ be the third side. Then $b^2 = 10^2 + 9^2 - 2(10)(9) \cos 60^{\circ} = 100 + 81 - 180(0.5) = 181 - 90 = 91$.
Thus $b = \sqrt{91}$. However,the problem implies the sides are $a, b, c$ where $B=60^{\circ}$.
If $a=10, b=9$,then $9^2 = 10^2 + c^2 - 2(10)(c) \cos 60^{\circ} \implies 81 = 100 + c^2 - 10c \implies c^2 - 10c + 19 = 0$.
Solving for $c$: $c = \frac{10 \pm \sqrt{100 - 76}}{2} = 5 \pm \sqrt{6}$.
Case $1$: $c = 5 + \sqrt{6}$. Perimeter $= 10 + 9 + 5 + \sqrt{6} = 24 + \sqrt{6}$.
Case $2$: $c = 5 - \sqrt{6}$. Perimeter $= 10 + 9 + 5 - \sqrt{6} = 24 - \sqrt{6}$.
Given the options,$24+\sqrt{6}$ is a valid perimeter.

(B) Using the Law of Sines,we have $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Substituting the given values: $\frac{5}{3/4} = \frac{7}{\sin B}$.
$\frac{20}{3} = \frac{7}{\sin B} \implies \sin B = \frac{21}{20}$.
Since the value of $\sin B$ must be $\le 1$ and $\frac{21}{20} > 1$,no such angle $B$ exists.
Therefore,no triangle $ABC$ can be formed with the given parameters.

(A) Since $A, B, C$ are in $A.P.$,we have $2B = A + C$.
Given $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin A$ and $b = k \sin B$.
The expression becomes $\frac{\sin A}{\sin B} \sin 2B + \frac{\sin B}{\sin A} \sin 2A$.
Substituting $B = 60^{\circ}$,$\sin B = \frac{\sqrt{3}}{2}$ and $\sin 2B = \sin 120^{\circ} = \frac{\sqrt{3}}{2}$.
Expression $= \frac{\sin A}{\sqrt{3}/2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}/2}{\sin A} \cdot 2 \sin A \cos A = \sin A + \sqrt{3} \cos A$.
Since $A + C = 120^{\circ}$,$C = 120^{\circ} - A$.
Using $\sin A + \sqrt{3} \cos A = 2(\frac{1}{2} \sin A + \frac{\sqrt{3}}{2} \cos A) = 2 \sin(A + 60^{\circ})$.
Since $A + 60^{\circ} = A + B = 180^{\circ} - C$,this simplifies to $2 \sin(180^{\circ} - C) = 2 \sin C$.
However,evaluating the expression $\frac{a}{b} \sin 2B + \frac{b}{a} \sin 2A$ specifically:
$= \frac{\sin A}{\sin B} (2 \sin B \cos B) + \frac{\sin B}{\sin A} (2 \sin A \cos A) = 2 \sin A \cos B + 2 \sin B \cos A = 2 \sin(A + B)$.
Since $A + B = 180^{\circ} - C$,$2 \sin(180^{\circ} - C) = 2 \sin C$.
Given $B = 60^{\circ}$,$A+C = 120^{\circ}$.
Actually,$2 \sin(A+B) = 2 \sin(180^{\circ} - C) = 2 \sin C$.
Wait,$2 \sin(A+B) = 2 \sin(120^{\circ}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$.

(C) Given the expression $E = \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)$.
Simplifying the terms,we get $E = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$.
This can be written as $E = \frac{(b+c)+a}{c} \times \frac{(b+c)-a}{b} = \frac{(b+c)^2 - a^2}{bc}$.
Expanding the numerator,we get $E = \frac{b^2+c^2+2bc-a^2}{bc} = \frac{b^2+c^2-a^2}{bc} + 2$.
Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $b^2+c^2-a^2 = 2bc \cos A$.
Substituting this into the expression,$E = \frac{2bc \cos A}{bc} + 2 = 2 \cos A + 2$.
Given $\angle A = 30^{\circ}$,we have $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
Thus,$E = 2 \left(\frac{\sqrt{3}}{2}\right) + 2 = \sqrt{3} + 2$.

(A) Let $AB = c$,$AC = b$,and $BC = a$. Let $\angle BAD = \alpha$ and $\angle CAD = \beta$.
In $\triangle ABD$,by the sine rule: $\frac{BD}{\sin \alpha} = \frac{c}{\sin \angle ADB} \implies \sin \alpha = \frac{BD \sin \angle ADB}{c}$.
In $\triangle ACD$,by the sine rule: $\frac{CD}{\sin \beta} = \frac{b}{\sin \angle ADC} \implies \sin \beta = \frac{CD \sin \angle ADC}{c}$.
Since $\angle ADB + \angle ADC = \pi$,$\sin \angle ADB = \sin \angle ADC$.
Thus,$\frac{\sin \alpha}{\sin \beta} = \frac{BD}{CD} \cdot \frac{b}{c}$.
Given $BD:CD = 1:3$,so $BD/CD = 1/3$.
By the sine rule in $\triangle ABC$,$\frac{b}{\sin B} = \frac{c}{\sin C} \implies \frac{b}{c} = \frac{\sin(\pi/3)}{\sin(\pi/4)} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.
Therefore,$\frac{\sin \alpha}{\sin \beta} = \frac{1}{3} \cdot \frac{\sqrt{3}}{\sqrt{2}} = \frac{1}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{6}}$.

(B) The angles of triangle $ABC$ are in the ratio $2:3:7$. Let the common multiple be $x$.
$\angle A = 2x, \angle B = 3x, \angle C = 7x$.
Since the sum of the angles of a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ} \implies 12x = 180^{\circ} \implies x = 15^{\circ}$.
Thus,$\angle A = 30^{\circ}, \angle B = 45^{\circ}, \angle C = 105^{\circ}$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Substituting the values: $\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 45^{\circ}} = \frac{c}{\sin 105^{\circ}}$.
We know $\sin 30^{\circ} = \frac{1}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,and $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Therefore,$\frac{a}{1/2} = \frac{b}{1/\sqrt{2}} = \frac{c}{(\sqrt{3}+1)/(2\sqrt{2})}$.
Multiplying by $\frac{1}{2}$,we get $\frac{a}{\sqrt{2}} = \frac{b}{2} = \frac{c}{\sqrt{3}+1}$.
Thus,the ratio $a:b:c = \sqrt{2}: 2: (\sqrt{3}+1)$.

(C) Given that: $m \angle A = 45^{\circ}, m \angle B = 75^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,$m \angle C = 180^{\circ} - (45^{\circ} + 75^{\circ}) = 60^{\circ}$.
Using the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k \sin 45^{\circ} = \frac{k}{\sqrt{2}}$,$b = k \sin 75^{\circ} = k \sin(45^{\circ} + 30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,and $c = k \sin 60^{\circ} = \frac{k\sqrt{3}}{2}$.
We need to find $a + c\sqrt{2}$:
$a + c\sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{2} \cdot \sqrt{2} = \frac{k}{\sqrt{2}} + \frac{k\sqrt{3}}{\sqrt{2}} = \frac{k(1 + \sqrt{3})}{\sqrt{2}}$.
From the expression for $b$,we have $b = \frac{k(\sqrt{3} + 1)}{2\sqrt{2}}$,which implies $2b = \frac{k(\sqrt{3} + 1)}{\sqrt{2}}$.
Therefore,$a + c\sqrt{2} = 2b$.

(C) Given that the angles $A, B, C$ are in $A$.$P$.,we have $A+C = 2B$. Since $A+B+C = 180^{\circ}$,it follows that $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we have $\frac{b}{c} = \frac{\sin B}{\sin C}$.
Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$.
This simplifies to $\sin C = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus,$C = 45^{\circ}$.
Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

(D) Let the angles of the triangle be $5x, x, 6x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $5x + x + 6x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.
The angles are $75^{\circ}, 15^{\circ}, 90^{\circ}$.
Let the sides opposite to these angles be $a, b, c$ respectively. By the Sine Rule,$\frac{a}{\sin 75^{\circ}} = \frac{b}{\sin 15^{\circ}} = \frac{c}{\sin 90^{\circ}} = k$.
The smallest side is $b$ (opposite to $15^{\circ}$) and the greatest side is $c$ (opposite to $90^{\circ}$).
The ratio of the smallest side to the greatest side is $\frac{b}{c} = \frac{\sin 15^{\circ}}{\sin 90^{\circ}}$.
We know $\sin 15^{\circ} = \sin(45^{\circ}-30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Since $\sin 90^{\circ} = 1$,the ratio is $\frac{\sqrt{3}-1}{2\sqrt{2}} : 1 = \sqrt{3}-1 : 2\sqrt{2}$.

(C) We know that in any triangle $ABC$,$A+B+C = \pi$.
Therefore,$A+C = \pi - B$.
Substituting this into the expression:
$2 ac \sin \left(\frac{A-B+C}{2}\right) = 2 ac \sin \left(\frac{(A+C)-B}{2}\right)$
$= 2 ac \sin \left(\frac{(\pi-B)-B}{2}\right)$
$= 2 ac \sin \left(\frac{\pi-2B}{2}\right)$
$= 2 ac \sin \left(\frac{\pi}{2} - B\right)$
$= 2 ac \cos B$
Using the cosine rule,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting this value:
$= 2 ac \left(\frac{a^2+c^2-b^2}{2ac}\right)$
$= a^2+c^2-b^2$.

(B) Let $a, b, c$ be the side lengths opposite to vertices $A, B, C$ respectively. Here,$a = 3$,$b = 4$,and $c = \sqrt{23}$.
Using the cotangent rule,$\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,$\cot B = \frac{a^2+c^2-b^2}{4\Delta}$,and $\cot C = \frac{a^2+b^2-c^2}{4\Delta}$,where $\Delta$ is the area of the triangle.
Then,$\frac{\cot A+\cot C}{\cot B} = \frac{\frac{b^2+c^2-a^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta}}{\frac{a^2+c^2-b^2}{4\Delta}} = \frac{b^2+c^2-a^2+a^2+b^2-c^2}{a^2+c^2-b^2} = \frac{2b^2}{a^2+c^2-b^2}$.
Substituting the values: $a^2 = 9$,$b^2 = 16$,$c^2 = 23$.
$\frac{2(16)}{9+23-16} = \frac{32}{16} = 2$.

(D) Let $a = \sqrt{3}+1$,$b = \sqrt{3}-1$,and $C = 60^{\circ}$.
Using the Law of Cosines:
$c^2 = a^2 + b^2 - 2ab \cos C$
$c^2 = (\sqrt{3}+1)^2 + (\sqrt{3}-1)^2 - 2(\sqrt{3}+1)(\sqrt{3}-1) \cos 60^{\circ}$
$c^2 = (3+1+2\sqrt{3}) + (3+1-2\sqrt{3}) - 2(3-1) \times \frac{1}{2}$
$c^2 = 8 - 2 = 6 \implies c = \sqrt{6}$.
Using the Law of Tangents:
$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$
$\tan\left(\frac{A-B}{2}\right) = \frac{(\sqrt{3}+1) - (\sqrt{3}-1)}{(\sqrt{3}+1) + (\sqrt{3}-1)} \cot(30^{\circ})$
$\tan\left(\frac{A-B}{2}\right) = \frac{2}{2\sqrt{3}} \times \sqrt{3} = 1$.
Therefore,$\frac{A-B}{2} = 45^{\circ} \implies A-B = 90^{\circ}$.

(B) Using the sine rule,we have:
$\frac{\sin A}{a} = \frac{\sin B}{b}$
Substituting the given values:
$\frac{\sin 60^{\circ}}{6} = \frac{\sin B}{8}$
Since $B = \sin^{-1} x$,we have $\sin B = x$.
$\frac{\sqrt{3}/2}{6} = \frac{x}{8}$
$\frac{\sqrt{3}}{12} = \frac{x}{8}$
$x = \frac{8\sqrt{3}}{12} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$

(D) By the cosine rule,we have:
$a^2 = b^2 + c^2 - 2bc \cos A$
$a^2 = (\sqrt{3})^2 + (1)^2 - 2(\sqrt{3})(1) \cos(30^{\circ})$
$a^2 = 3 + 1 - 2\sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$
$a^2 = 4 - 3 = 1$
$\therefore a = 1$
Since $b = \sqrt{3} \approx 1.732$ is the largest side,the largest angle is $\angle B$.
Using the cosine rule for $\angle B$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$
$\therefore B = 120^{\circ}$

(D) Let the angles be $x, 2x, 3x$.
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 of the triangle are $A = 30^{\circ}, B = 60^{\circ}, C = 90^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Substituting the values,$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$.
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$.
Multiplying by $1/2$,we get $a : b : c = \sin 30^{\circ} : \sin 60^{\circ} : \sin 90^{\circ} = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$.
Multiplying throughout by $2$,we get $a : b : c = 1 : \sqrt{3} : 2$.

(D) Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Then $\sin A = \frac{a}{k}$,$\sin B = \frac{b}{k}$,and $\sin C = \frac{c}{k}$.
The perimeter of the triangle is $a+b+c$.
The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
According to the problem,$a+b+c = 6 \times \left( \frac{\sin A + \sin B + \sin C}{3} \right)$.
Substituting the values,we get $a+b+c = 2(\sin A + \sin B + \sin C) = 2 \left( \frac{a+b+c}{k} \right)$.
Thus,$1 = \frac{2}{k}$,which implies $k = 2$.
Since $\sin A = \frac{a}{k}$ and $a=1$,we have $\sin A = \frac{1}{2}$.
Therefore,$A = \frac{\pi^c}{6}$.

(B) Given that the angles $A, B, C$ of $\triangle ABC$ are in $A$.$P$.,we have $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$ and $A + C = 120^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = k$,so $a = k \sin A$ and $c = k \sin C$.
Substituting these into the expression:
$\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A = \frac{k \sin A}{k \sin C} (2 \sin C \cos C) + \frac{k \sin C}{k \sin A} (2 \sin A \cos A)$
$= 2 \sin A \cos C + 2 \sin C \cos A$
$= 2(\sin A \cos C + \cos A \sin C)$
$= 2 \sin(A + C)$
$= 2 \sin(120^{\circ}) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

(B) We know that in $\triangle ABC$,$A+B+C = \pi$,so $A+B = \pi-C$.
Substituting this into the expression:
$2ab \sin \frac{1}{2}(A+B-C) = 2ab \sin \frac{1}{2}((\pi-C)-C)$
$= 2ab \sin \frac{1}{2}(\pi-2C) = 2ab \sin (\frac{\pi}{2}-C)$
$= 2ab \cos C$
Using the cosine rule,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Therefore,$2ab \cos C = 2ab \left(\frac{a^2+b^2-c^2}{2ab}\right) = a^2+b^2-c^2$.

(A) We know the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$.
This implies $a \sin B = b \sin A$ ... $(1)$.
We are given the condition: $a \cos B = b \cos A$ ... $(2)$.
Dividing equation $(1)$ by equation $(2)$,we get:
$\frac{a \sin B}{a \cos B} = \frac{b \sin A}{b \cos A} \Rightarrow \tan B = \tan A$.
Since $A$ and $B$ are angles of a triangle,$A = B$.
Therefore,the triangle is an isosceles triangle.

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Therefore,$b = k \sin B$ and $c = k \sin C$.
Substituting these into the expression:
$\frac{b \sin B - c \sin C}{\sin (B - C)} = \frac{(k \sin B) \sin B - (k \sin C) \sin C}{\sin (B - C)}$
$= \frac{k(\sin^2 B - \sin^2 C)}{\sin (B - C)}$
Using the identity $\sin^2 B - \sin^2 C = \sin(B - C) \sin(B + C)$:
$= \frac{k \sin(B - C) \sin(B + C)}{\sin (B - C)}$
$= k \sin(B + C)$
Since $A + B + C = \pi$,we have $\sin(B + C) = \sin(\pi - A) = \sin A$.
Thus,the expression becomes $k \sin A = a$.

(C) Given $\frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)}$.
Applying cross-multiplication: $\sin A \sin (B-C) = \sin C \sin (A-B)$.
Using the expansion $\sin(x-y) = \sin x \cos y - \cos x \sin y$:
$\sin A (\sin B \cos C - \cos B \sin C) = \sin C (\sin A \cos B - \cos A \sin B)$.
$\sin A \sin B \cos C - \sin A \cos B \sin C = \sin C \sin A \cos B - \sin C \cos A \sin B$.
Rearranging terms: $\sin A \sin B \cos C + \sin C \cos A \sin B = 2 \sin A \cos B \sin C$.
$\sin B (\sin A \cos C + \cos A \sin C) = 2 \sin A \cos B \sin C$.
Since $\sin A \cos C + \cos A \sin C = \sin(A+C) = \sin(\pi - B) = \sin B$,we have:
$\sin B \cdot \sin B = 2 \sin A \cos B \sin C$.
$\sin^2 B = 2 \sin A \sin C \cos B$.
Dividing by $\sin A \sin C \sin B$: $\frac{\sin B}{\sin A \sin C} = \frac{2 \cos B}{\sin B} = 2 \cot B$.
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Substituting these: $\frac{(b/2R)^2}{(a/2R)(c/2R)} = 2 \cos B \Rightarrow \frac{b^2}{ac} = 2 \cos B$.
Using the cosine rule $\cos B = \frac{a^2+c^2-b^2}{2ac}$:
$\frac{b^2}{ac} = 2 \left( \frac{a^2+c^2-b^2}{2ac} \right) = \frac{a^2+c^2-b^2}{ac}$.
$b^2 = a^2+c^2-b^2 \Rightarrow 2b^2 = a^2+c^2$.
Thus,$a^2, b^2, c^2$ are in $AP$.

(A) Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$,the expression becomes:
$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \left(\frac{1}{a^2} - \frac{1}{b^2}\right) - 2\left(\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2}\right)$
From the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = k$,which implies $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{1}{k}$.
Therefore,$\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = \frac{1}{k^2}$.
Substituting this into the expression:
$\left(\frac{1}{a^2} - \frac{1}{b^2}\right) - 2\left(\frac{1}{k^2} - \frac{1}{k^2}\right) = \frac{1}{a^2} - \frac{1}{b^2}$
Given $a=2$ and $b=3$:
$\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$

(B) Let the angles of $\triangle ABC$ be $A = \frac{\pi}{4} = 45^{\circ}$,$B = \frac{\pi}{3} = 60^{\circ}$,and $C$.
The sum of angles in a triangle is $\pi$ radians $(180^{\circ})$.
So,$C = 180^{\circ} - (45^{\circ} + 60^{\circ}) = 75^{\circ}$.
The angles are $45^{\circ}, 60^{\circ}, 75^{\circ}$.
The smallest angle is $45^{\circ}$ and the greatest angle is $75^{\circ}$.
By the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C}$,where $a$ is the smallest side and $c$ is the greatest side.
The ratio of the smallest side to the greatest side is $\frac{a}{c} = \frac{\sin 45^{\circ}}{\sin 75^{\circ}}$.
We know $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$.
Thus,$\frac{a}{c} = \frac{1/\sqrt{2}}{(\sqrt{3}+1)/(2\sqrt{2})} = \frac{2}{\sqrt{3}+1}$.
Rationalizing the denominator: $\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(\sqrt{3}-1)}{3-1} = \sqrt{3}-1$.
Therefore,the ratio is $(\sqrt{3}-1) : 1$.

(C) By the Sine Rule,we have: $\frac{\sin A}{a} = \frac{\sin C}{c}$
Substituting the given values: $\frac{\sin A}{3} = \frac{(2/3)}{2}$
$\frac{\sin A}{3} = \frac{1}{3}$
$\sin A = 1$
Therefore,$A = 90^{\circ} = \frac{\pi}{2}$.

(B) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Given the condition $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,we substitute the values of $a, b, c$:
$\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$\cot A = \cot B = \cot C$ implies $A = B = C$.
Therefore,the triangle is an equilateral triangle.

(D) Given $a=\sqrt{3}+1$,$b=\sqrt{3}-1$,$m \angle C=60^{\circ}$.
Using the tangent rule (Napier's Analogy): $\tan \left( \frac{A-B}{2} \right) = \frac{a-b}{a+b} \cot \left( \frac{C}{2} \right)$.
Substitute the values: $\frac{a-b}{a+b} = \frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
$\tan \left( \frac{A-B}{2} \right) = \frac{1}{\sqrt{3}} \cot \left( \frac{60^{\circ}}{2} \right) = \frac{1}{\sqrt{3}} \cot(30^{\circ}) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$.
$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$.

(C) Given the equation: $2 \cos C = \sin B \cdot \operatorname{cosec} A$
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $2 \cos C = \frac{\sin B}{\sin A}$.
Using the Sine Rule,$\frac{\sin B}{\sin A} = \frac{b}{a}$.
Thus,$2 \cos C = \frac{b}{a}$.
Using the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting this into the equation: $2 \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \frac{b}{a}$.
$\frac{a^2 + b^2 - c^2}{ab} = \frac{b}{a}$.
Multiplying both sides by $ab$,we get $a^2 + b^2 - c^2 = b^2$.
$a^2 - c^2 = 0 \Rightarrow a^2 = c^2$.
Since $a$ and $c$ are side lengths,$a = c$.

(D) Given that $A, B, C$ are in $A$.$P$.,we have $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we substitute $A + C = 2B$ to get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Using the Sine Rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,we have $\sin C = \frac{c}{b} \sin B$.
Given $b:c = \sqrt{3}:\sqrt{2}$,we have $\frac{c}{b} = \frac{\sqrt{2}}{\sqrt{3}}$.
Thus,$\sin C = \frac{\sqrt{2}}{\sqrt{3}} \times \sin 60^{\circ} = \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$C = 45^{\circ}$.
Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

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

(A) Key Idea: Use the sine rule,i.e.,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Given,$\sin B \sin C = \frac{bc}{a^2}$.
From the sine rule,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation:
$\left(\frac{b}{2R}\right) \left(\frac{c}{2R}\right) = \frac{bc}{a^2}$
$\Rightarrow \frac{bc}{4R^2} = \frac{bc}{a^2}$
$\Rightarrow 4R^2 = a^2$
$\Rightarrow a = 2R$.
Since $\frac{a}{\sin A} = 2R$,we have $\frac{2R}{\sin A} = 2R$,which implies $\sin A = 1$.
Therefore,$A = 90^{\circ}$.
Thus,the triangle is a right-angled triangle.

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

(A) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$.
Then $b+c = 11k$,$c+a = 12k$,and $a+b = 13k$.
Adding these equations,we get $2(a+b+c) = 36k$,so $a+b+c = 18k$.
Subtracting the equations from $a+b+c = 18k$:
$a = (a+b+c) - (b+c) = 18k - 11k = 7k$.
$b = (a+b+c) - (c+a) = 18k - 12k = 6k$.
$c = (a+b+c) - (a+b) = 18k - 13k = 5k$.
Using the cosine rule:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{(6k)^2+(5k)^2-(7k)^2}{2(6k)(5k)} = \frac{36+25-49}{60} = \frac{12}{60} = \frac{1}{5}$.
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{(7k)^2+(5k)^2-(6k)^2}{2(7k)(5k)} = \frac{49+25-36}{70} = \frac{38}{70} = \frac{19}{35}$.
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{(7k)^2+(6k)^2-(5k)^2}{2(7k)(6k)} = \frac{49+36-25}{84} = \frac{60}{84} = \frac{5}{7}$.
Now,$\cos A : \cos B : \cos C = \frac{1}{5} : \frac{19}{35} : \frac{5}{7}$.
Multiplying by $35$,we get $7 : 19 : 25$.

(B) Let the sides of the triangle be $a = 6+\sqrt{12} = 6+2\sqrt{3}$,$b = \sqrt{48} = 4\sqrt{3}$,and $c = \sqrt{24} = 2\sqrt{6}$.
Note that $a \approx 6 + 3.464 = 9.464$,$b \approx 4 \times 1.732 = 6.928$,and $c \approx 2 \times 2.449 = 4.898$.
The smallest side is $c = 2\sqrt{6}$. The smallest angle is opposite to the smallest side,so we need to find angle $C$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$a^2 = (6+2\sqrt{3})^2 = 36 + 12 + 24\sqrt{3} = 48 + 24\sqrt{3}$.
$b^2 = (4\sqrt{3})^2 = 48$.
$c^2 = (2\sqrt{6})^2 = 24$.
$\cos C = \frac{48 + 24\sqrt{3} + 48 - 24}{2(6+2\sqrt{3})(4\sqrt{3})} = \frac{72 + 24\sqrt{3}}{8\sqrt{3}(6+2\sqrt{3})} = \frac{24(3 + \sqrt{3})}{8\sqrt{3} \times 2(3 + \sqrt{3})} = \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos C = \frac{\sqrt{3}}{2}$,we have $C = \frac{\pi}{6}$.