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

(B) We know that $r = (s-a) \tan \frac{A}{2} = (s-b) \tan \frac{B}{2} = (s-c) \tan \frac{C}{2}$ and $r_1 = s \tan \frac{A}{2}$.
Given $\frac{r}{r_1} = \frac{1}{2}$,we have $\frac{(s-a) \tan \frac{A}{2}}{s \tan \frac{A}{2}} = \frac{1}{2}$,which implies $\frac{s-a}{s} = \frac{1}{2}$,so $2s - 2a = s$,or $s = 2a$.
Since $2s = a+b+c$,we have $a+b+c = 4a$,which means $b+c = 3a$.
We use the identity $\tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} + \tan \frac{A}{2} \tan \frac{B}{2} = 1$.
This can be written as $\tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 1 - \tan \frac{B}{2} \tan \frac{C}{2}$.
Using the formula $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$,we get $\tan \frac{B}{2} \tan \frac{C}{2} = \frac{s-a}{s}$.
Since $s = 2a$,$\frac{s-a}{s} = \frac{2a-a}{2a} = \frac{1}{2}$.
Thus,$\tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$4 \tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 4 \times \frac{1}{2} = 2$.

(D) Given that $a=7, b=8$ and $\cos C=\frac{2}{7}$.
Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting the values: $\frac{2}{7} = \frac{7^2+8^2-c^2}{2 \times 7 \times 8}$.
$\frac{2}{7} = \frac{49+64-c^2}{112}$.
Multiply both sides by $112$: $\frac{2 \times 112}{7} = 113 - c^2$.
$2 \times 16 = 113 - c^2$.
$32 = 113 - c^2$.
$c^2 = 113 - 32 = 81$.
Therefore,$c = \sqrt{81} = 9$.

(D) Given expression: $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right) = \left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$
$= \frac{(b+c)^2-a^2}{bc} = \frac{b^2+c^2-a^2+2bc}{bc}$
$= \frac{b^2+c^2-a^2}{bc} + 2 = 2\left(\frac{b^2+c^2-a^2}{2bc}\right) + 2$
$= 2\cos A + 2$
Since $A+B+C = 180^{\circ}$ and $B+C = 72^{\circ}$,we have $A = 180^{\circ} - 72^{\circ} = 108^{\circ}$.
So,the expression becomes $2\cos 108^{\circ} + 2 = 2\cos(90^{\circ} + 18^{\circ}) + 2$
$= -2\sin 18^{\circ} + 2 = -2\left(\frac{\sqrt{5}-1}{4}\right) + 2$
$= \frac{1-\sqrt{5}}{2} + 2 = \frac{1-\sqrt{5}+4}{2} = \frac{5-\sqrt{5}}{2}$.

(B) Using the projection formula,we know that $b = c \cos A + a \cos C$ and $c = b \cos A + a \cos B$.
Substituting these into the expression:
$\frac{b - c \cos A}{c - b \cos A} = \frac{(c \cos A + a \cos C) - c \cos A}{(b \cos A + a \cos B) - b \cos A}$
$= \frac{a \cos C}{a \cos B}$
$= \frac{\cos C}{\cos B}$

(B) Let the angles of $\triangle ABC$ be $(A-d), A, (A+d)$.
Since the sum of angles in a triangle is $180^{\circ}$,we have:
$(A-d) + A + (A+d) = 180^{\circ}$
$3A = 180^{\circ} \Rightarrow A = 60^{\circ}$.
Using the cosine rule for angle $A$:
$\cos A = \frac{b^2+c^2-a^2}{2bc}$
$\cos 60^{\circ} = \frac{b^2+c^2-a^2}{2bc}$
$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$
$bc = b^2+c^2-a^2$
$a^2 = b^2+c^2-bc$.
Thus,the correct relation is $a^2 = b^2+c^2-bc$.

(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$
Multiply both sides by $(a+b+c)$:
$\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3$
$1 + \frac{c}{a+b} + 1 + \frac{b}{c+a} = 3$
$\frac{c}{a+b} + \frac{b}{c+a} = 1$
$c(c+a) + b(a+b) = (a+b)(c+a)$
$c^2 + ac + ab + b^2 = ac + a^2 + bc + ab$
$b^2 + c^2 - a^2 = bc$
Using the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{bc}{2bc} = \frac{1}{2}$
Since $\cos A = \frac{1}{2}$,$A = 60^{\circ}$
Therefore,$\sin A = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

(C) Given: $a^2-c^2=b^2-bc$ $\Rightarrow a^2-c^2=b^2-bc$ $\Rightarrow b^2+c^2-a^2=bc$.
Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}$.
Thus,$A = 60^{\circ}$.
Using the sine rule,$\frac{a}{\sin A} = 2R$ $\Rightarrow \frac{a}{\sin 60^{\circ}} = 2 \times \frac{1}{\sqrt{3}}$ $\Rightarrow a = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = 1$.
Given $\sqrt{2}a = 2b-c$,substitute $a=1$: $\sqrt{2} = 2b-c \Rightarrow c = 2b-\sqrt{2}$.
Substitute $c$ into $a^2-c^2=b^2-bc$: $1^2 - (2b-\sqrt{2})^2 = b^2 - b(2b-\sqrt{2})$.
$1 - (4b^2 - 4\sqrt{2}b + 2) = b^2 - 2b^2 + \sqrt{2}b$.
$1 - 4b^2 + 4\sqrt{2}b - 2 = -b^2 + \sqrt{2}b$.
$-3b^2 + 3\sqrt{2}b - 1 = 0 \Rightarrow 3b^2 - 3\sqrt{2}b + 1 = 0$.
Using the quadratic formula $b = \frac{-(-3\sqrt{2}) \pm \sqrt{(-3\sqrt{2})^2 - 4(3)(1)}}{2(3)} = \frac{3\sqrt{2} \pm \sqrt{18-12}}{6} = \frac{3\sqrt{2} \pm \sqrt{6}}{6} = \frac{\sqrt{2}}{2} \pm \frac{\sqrt{6}}{6} = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{6} = \frac{3\sqrt{2}+\sqrt{6}}{6} = \frac{\sqrt{3}+1}{\sqrt{6}}$.

(A) Let the sides of the triangle be $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$.
Since $c$ is the largest side,the largest angle $\theta$ is opposite to side $c$.
Using the Law of Cosines: $\cos \theta = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos \theta = \frac{(\sqrt{3}k)^2 + (\sqrt{5}k)^2 - (\sqrt{8+\sqrt{15}}k)^2}{2(\sqrt{3}k)(\sqrt{5}k)}$.
$\cos \theta = \frac{3k^2 + 5k^2 - (8+\sqrt{15})k^2}{2\sqrt{15}k^2}$.
$\cos \theta = \frac{8k^2 - 8k^2 - \sqrt{15}k^2}{2\sqrt{15}k^2} = \frac{-\sqrt{15}k^2}{2\sqrt{15}k^2} = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2\pi}{3}$.

(B) Given the expression: $b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2}$
Using the identity $2 \cos^2 \theta = 1 + \cos(2\theta)$,we have:
$= \frac{1}{2} [b(1 + \cos C) + c(1 + \cos B)]$
$= \frac{1}{2} [b + c + b \cos C + c \cos B]$
By the projection rule,$a = b \cos C + c \cos B$.
Substituting this into the expression:
$= \frac{1}{2} [b + c + a] = \frac{1}{2} (a + b + c)$
Since the perimeter $a + b + c = 50 \text{ cm}$:
$= \frac{50}{2} = 25 \text{ cm}$.
Thus,option $B$ is correct.

(C) From the sine rule: $\frac{\alpha+1}{\sin A} = \frac{\alpha-1}{\sin B} = \frac{\alpha}{\sin C}$.
Given $A = 2B$,so $\frac{\alpha+1}{\sin 2B} = \frac{\alpha-1}{\sin B}$.
Using $\sin 2B = 2 \sin B \cos B$,we get $\frac{\alpha+1}{2 \sin B \cos B} = \frac{\alpha-1}{\sin B}$,which simplifies to $\cos B = \frac{\alpha+1}{2(\alpha-1)} \dots(1)$.
From the cosine rule: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(\alpha+1)^2 + \alpha^2 - (\alpha-1)^2}{2(\alpha+1)(\alpha)}$.
Simplifying the numerator: $(\alpha^2 + 2\alpha + 1) + \alpha^2 - (\alpha^2 - 2\alpha + 1) = \alpha^2 + 4\alpha$.
So,$\cos B = \frac{\alpha^2 + 4\alpha}{2\alpha(\alpha+1)} = \frac{\alpha+4}{2(\alpha+1)} \dots(2)$.
Equating $(1)$ and $(2)$: $\frac{\alpha+1}{2(\alpha-1)} = \frac{\alpha+4}{2(\alpha+1)}$.
$(\alpha+1)^2 = (\alpha-1)(\alpha+4) \Rightarrow \alpha^2 + 2\alpha + 1 = \alpha^2 + 3\alpha - 4$.
Solving for $\alpha$,we get $\alpha = 5$.

(A) Given,$a^4+b^4+c^4=2b^2c^2+2a^2b^2$.
Rearranging the terms,we get $a^4+b^4+c^4-2b^2c^2-2a^2b^2=0$.
Adding $2a^2c^2$ to both sides,we have $a^4+c^4+b^4-2a^2b^2-2b^2c^2+2a^2c^2 = 2a^2c^2$.
This simplifies to $(a^2+c^2-b^2)^2 = 2a^2c^2$.
Dividing by $4a^2c^2$,we get $\frac{(a^2+c^2-b^2)^2}{4a^2c^2} = \frac{2a^2c^2}{4a^2c^2} = \frac{1}{2}$.
Since $\cos B = \frac{a^2+c^2-b^2}{2ac}$,we have $\cos^2 B = \frac{1}{2}$.
Thus,$\cos B = \pm \frac{1}{\sqrt{2}}$.
Therefore,$B = \frac{\pi}{4}$ or $B = \frac{3\pi}{4}$.

(C) In $\triangle ACD$,since $AD \perp AC$,$\angle DAC = 90^\circ$. Thus,$\cos C = \frac{AC}{CD} = \frac{b}{a/2} = \frac{2b}{a}$.
In $\triangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Equating the two expressions for $\cos C$:
$\frac{2b}{a} = \frac{a^2+b^2-c^2}{2ab}$
$4b^2 = a^2+b^2-c^2$
$3b^2 = a^2-c^2 \implies b^2 = \frac{a^2-c^2}{3}$.
Now,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.
$\cos A \cos C = \left(\frac{b^2+c^2-a^2}{2bc}\right) \left(\frac{2b}{a}\right) = \frac{b^2+c^2-a^2}{ac}$.
Substituting $b^2 = \frac{a^2-c^2}{3}$:
$\cos A \cos C = \frac{\frac{a^2-c^2}{3} + c^2 - a^2}{ac} = \frac{a^2-c^2+3c^2-3a^2}{3ac} = \frac{2c^2-2a^2}{3ac} = \frac{2(c^2-a^2)}{3ac}$.

(D) Let the angles of $\triangle ABC$ be $A-d, A, A+d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.
Thus,the angles are $60^{\circ}-d, 60^{\circ}, 60^{\circ}+d$.
Using the Law of Cosines for the angle $60^{\circ}$:
$\cos 60^{\circ} = \frac{a^2+c^2-b^2}{2ac} = \frac{1}{2}$
$a^2+c^2-b^2 = ac$
$c^2 - ac + (a^2-b^2) = 0$
Solving this quadratic equation for $c$ using the quadratic formula:
$c = \frac{a \pm \sqrt{a^2 - 4(a^2-b^2)}}{2}$
$c = \frac{a \pm \sqrt{4b^2-3a^2}}{2}$

(C) Given the equation: $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$
Taking $LCM$ on the left side: $\frac{(b+c) + (a+c)}{(a+c)(b+c)} = \frac{3}{a+b+c}$
$\Rightarrow \frac{a+b+2c}{ab + ac + bc + c^2} = \frac{3}{a+b+c}$
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(ab + ac + bc + c^2)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3ab + 3ac + 3bc + 3c^2$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2$
$a^2 + b^2 - ab = c^2$
Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$
Comparing $a^2 + b^2 - ab = c^2$ with $a^2 + b^2 - 2ab \cos C = c^2$,we get:
$ab = 2ab \cos C$
$\cos C = \frac{1}{2}$
Therefore,$\angle C = 60^{\circ}$.

(A) Given: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$:
$a \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-a)}{bc} = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$
Since $2s = a + b + c$,we have $2s - a - c = b$:
$\frac{s}{b} (b) = \frac{3b}{2}$
$s = \frac{3b}{2}$
$2s = 3b$
$a + b + c = 3b$
$a + c = 2b$
Therefore,$a, b, c$ are in arithmetic progression.

(B) Using the projection formula or the cosine rule:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting these into the expression:
$a(b \cos C - c \cos B) = ab \cos C - ac \cos B$
$= ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right)$
$= \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2}$
$= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$
$= \frac{2b^2 - 2c^2}{2}$
$= b^2 - c^2$.

(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$
Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$
$\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
Simplifying by subtracting common terms from both sides: $a^2 + b^2 - c^2 = ab$
Using the Cosine Rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substituting $a^2 + b^2 - c^2 = ab$: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$
Since $\cos C = \frac{1}{2}$,we have $\angle C = 60^{\circ}$.

(B) We have,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
$= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$
$= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= (a^2 + b^2)(1) - 2ab \cos C$
Since $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,we have $a^2 + b^2 - 2ab \cos C = c^2$
Therefore,the expression is equal to $c^2$.

(B) We know the formula for the cotangent of half-angles in a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get: $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.
Given $3a = b + c$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.
Substituting $s = 2a$ into the expression: $\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.

(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$.
Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$:
$\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)} + \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)} = 0$
$\Rightarrow \frac{\sin A}{\sin(B+C)\sin(B-C)} + \frac{\sin C}{\sin(B+A)\sin(B-A)} = 0$
Since $A+B+C = \pi$,we have $\sin(B+C) = \sin A$ and $\sin(B+A) = \sin C$.
$\Rightarrow \frac{\sin A}{\sin A \sin(B-C)} + \frac{\sin C}{\sin C \sin(B-A)} = 0$
$\Rightarrow \frac{1}{\sin(B-C)} + \frac{1}{\sin(B-A)} = 0$
$\Rightarrow \sin(B-A) + \sin(B-C) = 0$
$\Rightarrow 2 \sin\left(\frac{2B-A-C}{2}\right) \cos\left(\frac{A-C}{2}\right) = 0$
Assuming $\cos\left(\frac{A-C}{2}\right) \neq 0$,we get $\sin\left(\frac{2B-(A+C)}{2}\right) = 0$.
Since $A+C = \pi - B$,we have $\frac{2B-(\pi-B)}{2} = 0$ $\Rightarrow 3B = \pi$ $\Rightarrow B = \frac{\pi}{3}$.

(B) Using the sine rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Consider the first term:
$\frac{\cos C+\cos A}{c+a} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(\sin C+\sin A)} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R \cdot 2 \sin \frac{C+A}{2} \cos \frac{C-A}{2}} = \frac{1}{2R} \cot \frac{C+A}{2} = \frac{1}{2R} \tan \frac{B}{2}$.
Consider the second term:
$\frac{\cos B}{b} = \frac{\cos B}{2R \sin B} = \frac{1}{2R} \cot B$.
Adding them together:
$\frac{1}{2R} \left( \tan \frac{B}{2} + \cot B \right) = \frac{1}{2R} \left( \tan \frac{B}{2} + \frac{1-\tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{2 \tan^2 \frac{B}{2} + 1 - \tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{1 + \tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \cdot \frac{1}{\sin B} = \frac{1}{b}$.

(A) The perimeter of $\triangle ABC$ is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
Given: $a+b+c = 6 \times \left(\frac{\sin A + \sin B + \sin C}{3}\right) = 2(\sin A + \sin B + \sin C)$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we have:
$2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
This implies $2R = 2$,so $R = 1$.
Given $BC = a = 1$,we use $a = 2R \sin A$:
$1 = 2(1) \sin A \implies \sin A = \frac{1}{2}$.
Since $A$ is an angle of a triangle,$A = 30^{\circ} = \frac{\pi}{6}$.

(B) Since $BC$ is the hypotenuse,the triangle is right-angled at $A$,therefore $\angle A = \frac{\pi}{2}$.
We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.
Thus,$r_2 + r_3 = \Delta \left( \frac{1}{s-b} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} \right) = \Delta \left( \frac{2s-b-c}{(s-b)(s-c)} \right)$.
Since $2s = a+b+c$,we have $2s-b-c = a$.
Also,$\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-b)(s-c) = \frac{\Delta^2}{s(s-a)}$.
Substituting these,$r_2 + r_3 = \Delta \left( \frac{a}{\Delta^2 / s(s-a)} \right) = \frac{as(s-a)}{\Delta}$.
Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{s(s-a)}{\Delta} = \cot \frac{A}{2}$.
Therefore,$r_2 + r_3 = a \cot \frac{A}{2} = a \cot \frac{\pi}{4} = a(1) = a$.

(D) Given $a=3, b=7, c=8$.
Using the Law of Tangents,$\tan \frac{C-A}{2} = \frac{c-a}{c+a} \cot \frac{B}{2}$.
Substituting this into the expression:
$\sin \frac{B}{2} \tan \frac{C-A}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} \right) \cot \frac{B}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} \right) \frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} = \left( \frac{c-a}{c+a} \right) \cos \frac{B}{2}$.
Now,calculate $\cos B$ using the Law of Cosines:
$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{3^2+8^2-7^2}{2 \times 3 \times 8} = \frac{9+64-49}{48} = \frac{24}{48} = \frac{1}{2}$.
Using the half-angle formula $\cos \frac{B}{2} = \sqrt{\frac{1+\cos B}{2}}$:
$\cos \frac{B}{2} = \sqrt{\frac{1+1/2}{2}} = \sqrt{\frac{3/2}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Finally,substitute the values:
$\left( \frac{8-3}{8+3} \right) \cos \frac{B}{2} = \frac{5}{11} \times \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{22}$.
Thus,option $(d)$ is correct.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{B}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{s^2}{(s-c)^2}} = \frac{s}{s-c}$.
Given $\frac{s}{s-c} = K$,we can write $K = \frac{s-c+c}{s-c} = 1 + \frac{c}{s-c}$.
Since $s = \frac{a+b+c}{2}$,we have $s-c = \frac{a+b-c}{2}$.
Thus,$K = 1 + \frac{c}{\frac{a+b-c}{2}} = 1 + \frac{2c}{a+b-c}$.
By the triangle inequality,$a+b > c$,so $a+b-c > 0$,which implies $K > 1$.
As $C \to 180^{\circ}$,$a+b-c \to 0^{+}$,so $K \to \infty$.
Therefore,the range of $K$ is $(1, \infty)$.

(D) Given,$4 \Delta = c^2 - (a - b)^2$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get:
$4 \Delta = (c - (a - b))(c + (a - b)) = (c - a + b)(c + a - b)$.
Dividing by $4$,we have:
$\Delta = \left(\frac{b + c - a}{2}\right) \left(\frac{a + c - b}{2}\right) = (s - a)(s - b)$,where $s$ is the semi-perimeter.
From Heron's formula,$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Thus,$\sqrt{s(s - a)(s - b)(s - c)} = (s - a)(s - b)$.
Squaring both sides: $s(s - a)(s - b)(s - c) = (s - a)^2(s - b)^2$.
$\frac{s(s - c)}{(s - a)(s - b)} = 1$.
We know that $\tan^2\left(\frac{C}{2}\right) = \frac{(s - a)(s - b)}{s(s - c)}$.
Therefore,$\tan^2\left(\frac{C}{2}\right) = 1$,which implies $\tan\left(\frac{C}{2}\right) = 1$.
$\frac{C}{2} = \frac{\pi}{4} \Rightarrow C = \frac{\pi}{2}$.
Hence,$\sin C = \sin\left(\frac{\pi}{2}\right) = 1$.

(B) We know that $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ and $\tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.
Multiplying these,we get $\tan \frac{B}{2} \tan \frac{C}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-a)^2}{s^2}} = \frac{s-a}{s}$.
Substituting $s = \frac{a+b+c}{2}$,we have $\frac{s-a}{s} = \frac{\frac{a+b+c}{2} - a}{\frac{a+b+c}{2}} = \frac{b+c-a}{b+c+a}$.
Given $b+c = 4a$,we substitute this into the expression:
$\frac{4a-a}{4a+a} = \frac{3a}{5a} = \frac{3}{5}$.

(C) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.
Multiplying these,we get $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a) \cdot s(s-c)}{(s-b)(s-c) \cdot (s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.
Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.
Thus,$\frac{s}{s-b} = \frac{2s}{2s-2b} = \frac{a+b+c}{a+b+c-2b} = \frac{(a+c)+b}{(a+c)-b}$.
Given $a+c=5b$,substituting this into the expression gives $\frac{5b+b}{5b-b} = \frac{6b}{4b} = \frac{3}{2}$.

(A) Given for triangle $ABC$: $c=9, s=10, \Delta=10\sqrt{2}$.
Using Napier's Analogy: $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}\cot\left(\frac{C}{2}\right)$.
We know $\cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{\Delta} = \frac{10(10-9)}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cdot \frac{1}{\sqrt{2}}$,which implies $\sqrt{2}\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b}$.
Now,substitute this into the expression $b\left[1+\sqrt{2}\tan\left(\frac{A-B}{2}\right)\right]$:
$= b\left[1 + \frac{a-b}{a+b}\right] = b\left[\frac{a+b+a-b}{a+b}\right] = b\left[\frac{2a}{a+b}\right]$.
Rearranging the terms: $a\left[\frac{2b}{a+b}\right] = a\left[\frac{(a+b)-(a-b)}{a+b}\right] = a\left[1 - \frac{a-b}{a+b}\right]$.
Substituting back $\frac{a-b}{a+b} = \sqrt{2}\tan\left(\frac{A-B}{2}\right)$,we get:
$= a\left[1 - \sqrt{2}\tan\left(\frac{A-B}{2}\right)\right]$.

(B) Let the sides of the triangle be $x-1, x, x+1$ where $x > 2$ (to ensure all sides are positive). The smallest side is $x-1$ and the largest side is $x+1$. Let the smallest angle be $\theta$ (opposite to side $x-1$) and the largest angle be $2\theta$ (opposite to side $x+1$).
Using the Law of Cosines for the angle $2\theta$ (opposite to side $x+1$):
$\cos(2\theta) = \frac{x^2 + (x-1)^2 - (x+1)^2}{2x(x-1)} = \frac{x^2 + x^2 - 2x + 1 - (x^2 + 2x + 1)}{2x(x-1)} = \frac{x^2 - 4x}{2x(x-1)} = \frac{x-4}{2(x-1)}$
Using the Law of Sines:
$\frac{x+1}{\sin(2\theta)} = \frac{x-1}{\sin(\theta)} \Rightarrow \frac{x+1}{2\sin(\theta)\cos(\theta)} = \frac{x-1}{\sin(\theta)} \Rightarrow \cos(\theta) = \frac{x+1}{2(x-1)}$
Since $\cos(2\theta) = 2\cos^2(\theta) - 1$,we substitute the expressions:
$\frac{x-4}{2(x-1)} = 2\left(\frac{x+1}{2(x-1)}\right)^2 - 1 = \frac{(x+1)^2}{2(x-1)^2} - 1 = \frac{x^2+2x+1 - 2(x^2-2x+1)}{2(x-1)^2} = \frac{-x^2+6x-1}{2(x-1)^2}$
Solving $\frac{x-4}{x-1} = \frac{-x^2+6x-1}{(x-1)^2} \Rightarrow (x-4)(x-1) = -x^2+6x-1 \Rightarrow x^2-5x+4 = -x^2+6x-1 \Rightarrow 2x^2-11x+5 = 0 \Rightarrow (2x-1)(x-5) = 0$. Since $x$ is a natural number and $x>2$,we get $x=5$.
The sides are $4, 5, 6$. The semi-perimeter $s = \frac{4+5+6}{2} = \frac{15}{2}$.
Area = $\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15}{2} \times (\frac{15}{2}-4) \times (\frac{15}{2}-5) \times (\frac{15}{2}-6)} = \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}} = \sqrt{\frac{1575}{16}} = \frac{15\sqrt{7}}{4}$.

(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
From this,we get $\frac{1}{p_1} = \frac{a}{2\Delta}$,$\frac{1}{p_2} = \frac{b}{2\Delta}$,and $\frac{1}{p_3} = \frac{c}{2\Delta}$.
Substituting these into the expression:
$\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$.
Since $2s = a+b+c$,we have $a+b = 2s-c$.
Therefore,$\frac{a+b-c}{2\Delta} = \frac{(2s-c)-c}{2\Delta} = \frac{2s-2c}{2\Delta} = \frac{s-c}{\Delta}$.

(D) Given,$r_1 = 2r_2 = 3r_3$.
Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,we have:
$\frac{\Delta}{s-a} = \frac{2\Delta}{s-b} = \frac{3\Delta}{s-c}$.
Dividing by $\Delta$,we get $\frac{1}{s-a} = \frac{2}{s-b} = \frac{3}{s-c} = k$ (let).
Then $s-a = \frac{1}{k}$,$s-b = \frac{2}{k}$,and $s-c = \frac{3}{k}$.
Adding these,$(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = \frac{1+2+3}{k} = \frac{6}{k}$,which implies $k = \frac{6}{s}$.
Now,$s-a = \frac{1}{6/s} = \frac{s}{6}$ $\Rightarrow 6s - 6a = s$ $\Rightarrow 5s = 6a$ $\Rightarrow a = \frac{5s}{6}$.
And $s-b = \frac{2}{6/s} = \frac{2s}{6} = \frac{s}{3}$ $\Rightarrow 3s - 3b = s$ $\Rightarrow 2s = 3b$ $\Rightarrow b = \frac{2s}{3} = \frac{4s}{6}$.
Therefore,$a : b = \frac{5s}{6} : \frac{4s}{6} = 5 : 4$.

(B) Given $a+3b=3c$,we have $a=3(c-b)$.
Using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s = \frac{a+b+c}{2}$.
$s-b = \frac{a+b+c}{2} - b = \frac{a-b+c}{2} = \frac{3(c-b)-b+c}{2} = \frac{4c-4b}{2} = 2(c-b)$.
$s-c = \frac{a+b+c}{2} - c = \frac{a+b-c}{2} = \frac{3(c-b)+b-c}{2} = \frac{2c-2b}{2} = c-b$.
Substituting these into the formula:
$\sin \frac{A}{2} = \sqrt{\frac{2(c-b)(c-b)}{bc}} = \sqrt{\frac{2(c-b)^2}{bc}} = (c-b) \sqrt{\frac{2}{bc}}$.
Since $c-b = \frac{a}{3}$,we get $\sin \frac{A}{2} = \frac{a}{3} \sqrt{\frac{2}{bc}}$.
Thus,option $B$ is correct.