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

(A) Let the cyclic quadrilateral be $ABCD$ with sides $AB = 2$,$BC = 5$,$CD = 3$,and $DA = d$. The angle $\angle ABC = 60^o$.
Since $ABCD$ is a cyclic quadrilateral,the opposite angles are supplementary. Therefore,$\angle ADC = 180^o - 60^o = 120^o$.
Using the law of cosines in $\triangle ABC$ for the diagonal $AC$:
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(60^o) = 2^2 + 5^2 - 2(2)(5)(0.5) = 4 + 25 - 10 = 19$.
Using the law of cosines in $\triangle ADC$ for the diagonal $AC$:
$AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos(120^o) = d^2 + 3^2 - 2(d)(3)(-0.5) = d^2 + 9 + 3d$.
Equating the two expressions for $AC^2$:
$d^2 + 3d + 9 = 19 \Rightarrow d^2 + 3d - 10 = 0$.
Factoring the quadratic equation:
$(d + 5)(d - 2) = 0$.
Since the side length $d$ must be positive,$d = 2$.

(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Given $a = 4$,$b = 3$,and $A = 60^\circ$,we have:
$\cos 60^\circ = \frac{3^2 + c^2 - 4^2}{2(3)(c)}$
$\frac{1}{2} = \frac{9 + c^2 - 16}{6c}$
$\frac{1}{2} = \frac{c^2 - 7}{6c}$
$6c = 2(c^2 - 7)$
$6c = 2c^2 - 14$
$2c^2 - 6c - 14 = 0$
Dividing by $2$,we get:
$c^2 - 3c - 7 = 0$.

(D) Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the given values: $\cos C = \frac{2^2 + 3^2 - 5^2}{2(2)(3)}$.
$\cos C = \frac{4 + 9 - 25}{12} = \frac{13 - 25}{12} = \frac{-12}{12} = -1$.
Since $\cos C = -1$,we have $C = 180^{\circ}$ or $\pi$ radians.
However,the sum of two sides of a triangle must be greater than the third side $(a + b > c)$.
Here,$2 + 3 = 5$,which means $a + b = c$.
This implies that the points $A, B, C$ are collinear and do not form a triangle.
Therefore,the correct option is $D$.

(B) Given $A = 30^{\circ}$,$a = 7$,and $b = 8$.
Calculate $b \sin A = 8 \sin 30^{\circ} = 8 \times 0.5 = 4$.
Since $b \sin A < a < b$ (i.e.,$4 < 7 < 8$),the triangle satisfies the condition for the ambiguous case.
Therefore,there are two possible values for angle $B$,which means there are two solutions.

(D) Given $b = 3, c = 4$ and $B = \frac{\pi}{3}$.
We calculate $c \sin B = 4 \sin \frac{\pi}{3} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$.
Since $\sqrt{3} \approx 1.732$,we have $2\sqrt{3} \approx 3.464$.
Comparing this with $b$,we find $c \sin B = 2\sqrt{3} > 3 = b$.
Since $b < c \sin B$,the side $b$ is too short to reach the third side of the triangle.
Therefore,no triangle can be constructed.

(C) Given that $a^2, b^2, c^2$ are in $A.P.$,we have $2b^2 = a^2 + c^2$.
Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these into the $A.P.$ condition: $2(2R \sin B)^2 = (2R \sin A)^2 + (2R \sin C)^2$.
This simplifies to $2 \sin^2 B = \sin^2 A + \sin^2 C$,which means $\sin^2 B - \sin^2 A = \sin^2 C - \sin^2 B$.
Using the identity $\sin^2 x - \sin^2 y = \sin(x+y)\sin(x-y)$,we get $\sin(B+A)\sin(B-A) = \sin(C+B)\sin(C-B)$.
Since $A+B+C = \pi$,we have $\sin(B+A) = \sin C$ and $\sin(C+B) = \sin A$.
Thus,$\sin C \sin(B-A) = \sin A \sin(C-B)$.
Expanding this,$\sin C(\sin B \cos A - \cos B \sin A) = \sin A(\sin C \cos B - \cos C \sin B)$.
Dividing both sides by $\sin A \sin B \sin C$,we obtain $\cot A - \cot B = \cot B - \cot C$.
Therefore,$\cot A, \cot B, \cot C$ are in $A.P.$

(D) Let the sides of the triangle be $a = n$,$b = n + 1$,and $c = n + 2$,where $n$ is a natural number.
Since $c > b > a$,the angle $C$ is the largest and $A$ is the smallest. Given $C = 2A$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$ and $\sin C = \frac{c}{2R}$.
Since $C = 2A$,$\sin C = \sin 2A = 2 \sin A \cos A$.
Substituting the values,$\frac{c}{2R} = 2 \left( \frac{a}{2R} \right) \cos A$,which simplifies to $\cos A = \frac{c}{2a}$.
Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Equating the two expressions for $\cos A$: $\frac{c}{2a} = \frac{b^2 + c^2 - a^2}{2bc} \implies c^2 b = a(b^2 + c^2 - a^2)$.
Substituting $a = n, b = n + 1, c = n + 2$:
$(n + 1)(n + 2)^2 = n((n + 1)^2 + (n + 2)^2 - n^2)$.
$(n + 1)(n^2 + 4n + 4) = n(n^2 + 2n + 1 + n^2 + 4n + 4 - n^2)$.
$(n + 1)(n^2 + 4n + 4) = n(n^2 + 6n + 5) = n(n + 1)(n + 5)$.
Since $n$ is a natural number,$n+1 \neq 0$,so $(n + 2)^2 = n(n + 5)$.
$n^2 + 4n + 4 = n^2 + 5n \implies n = 4$.
The sides are $4, 5, 6$.

(A) The perimeter of $\Delta ABC$ is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.
Given: $a + b + c = 6 \times \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)$.
Using the Sine Rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k = 2R$ (the diameter of the circumcircle).
Substituting these into the equation: $k(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.
Since $\sin A + \sin B + \sin C \neq 0$ for a triangle,we have $k = 2$.
Given $a = 1$,we use $a = k \sin A$:
$1 = 2 \sin A \implies \sin A = \frac{1}{2}$.
Since $A$ is an angle of a triangle,$A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$. Given the options,$A = \frac{\pi}{6}$.

(C) Let $BD = DE = EC = k$. Then $BE = 2k$,$DC = 2k$,$BC = 3k$.
Using the sine rule in $\Delta ABD$,$\frac{\sin x}{BD} = \frac{\sin B}{AD} \implies \frac{\sin x}{k} = \frac{\sin B}{AD} \implies \frac{AD}{\sin B} = \frac{k}{\sin x}$.
In $\Delta ABE$,$\frac{\sin(x + y)}{BE} = \frac{\sin B}{AE} \implies \frac{\sin(x + y)}{2k} = \frac{\sin B}{AE} \implies \frac{AE}{\sin B} = \frac{2k}{\sin(x + y)}$.
In $\Delta AEC$,$\frac{\sin z}{EC} = \frac{\sin C}{AE} \implies \frac{\sin z}{k} = \frac{\sin C}{AE} \implies \frac{AE}{\sin C} = \frac{k}{\sin z}$.
In $\Delta ADC$,$\frac{\sin(y + z)}{DC} = \frac{\sin C}{AD} \implies \frac{\sin(y + z)}{2k} = \frac{\sin C}{AD} \implies \frac{AD}{\sin C} = \frac{2k}{\sin(y + z)}$.
Now,consider the ratio $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z} = \left( \frac{\sin(x + y)}{\sin x} \right) \left( \frac{\sin(y + z)}{\sin z} \right)$.
From the sine rule ratios: $\frac{\sin(x + y)}{\sin x} = \frac{2k/AE}{k/AD} = \frac{2AD}{AE}$ and $\frac{\sin(y + z)}{\sin z} = \frac{2k/AD}{k/AE} = \frac{2AE}{AD}$.
Multiplying these: $\frac{2AD}{AE} \times \frac{2AE}{AD} = 4$.

(A) Given: $\cos A + 2\cos B + \cos C = 2$
Using the sum-to-product formula: $\cos A + \cos C = 2\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\right)$
Since $A+B+C = \pi$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cos\left(\frac{A+C}{2}\right) = \sin\left(\frac{B}{2}\right)$
Also,$2\cos B = 2(1 - 2\sin^2\frac{B}{2}) = 2 - 4\sin^2\frac{B}{2}$
Substituting these into the equation: $2\sin\left(\frac{B}{2}\right)\cos\left(\frac{A-C}{2}\right) + 2 - 4\sin^2\left(\frac{B}{2}\right) = 2$
$2\sin\left(\frac{B}{2}\right)\cos\left(\frac{A-C}{2}\right) = 4\sin^2\left(\frac{B}{2}\right)$
Dividing by $2\sin\left(\frac{B}{2}\right)$ (since $\sin\left(\frac{B}{2}\right) \neq 0$): $\cos\left(\frac{A-C}{2}\right) = 2\sin\left(\frac{B}{2}\right)$
Substitute $\sin\left(\frac{B}{2}\right) = \cos\left(\frac{A+C}{2}\right)$: $\cos\left(\frac{A-C}{2}\right) = 2\cos\left(\frac{A+C}{2}\right)$
Using $\cos(x-y) = \cos x \cos y + \sin x \sin y$ and $\cos(x+y) = \cos x \cos y - \sin x \sin y$:
$\cos\frac{A}{2}\cos\frac{C}{2} + \sin\frac{A}{2}\sin\frac{C}{2} = 2(\cos\frac{A}{2}\cos\frac{C}{2} - \sin\frac{A}{2}\sin\frac{C}{2})$
$3\sin\frac{A}{2}\sin\frac{C}{2} = \cos\frac{A}{2}\cos\frac{C}{2}$ $\Rightarrow \tan\frac{A}{2}\tan\frac{C}{2} = \frac{1}{3}$
Using $\tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this simplifies to $a+c = 2b$,hence $a, b, c$ are in $A.P.$

(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$.
Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = 180^o$,we have $\cos 3A + \cos 3B + \cos 3C = 1 - 4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2}$.
Equating this to $1$,we get $1 - 4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 1$.
This implies $4\sin \frac{3A}{2} \sin \frac{3B}{2} \sin \frac{3C}{2} = 0$.
Therefore,$\sin \frac{3A}{2} = 0$ or $\sin \frac{3B}{2} = 0$ or $\sin \frac{3C}{2} = 0$.
This means $\frac{3A}{2} = 180^o$ or $\frac{3B}{2} = 180^o$ or $\frac{3C}{2} = 180^o$.
Thus,$A = 120^o$ or $B = 120^o$ or $C = 120^o$.

(D) Given $AB = c$ and $BC = a$. Since $AB = 2BC$,we have $c = 2a$.
Using the Napier's Analogy,we know that $\tan \frac{C - A}{2} = \frac{c - a}{c + a} \cot \frac{B}{2}$.
Rearranging this,we get $\frac{\tan (B/2)}{\cot ((C - A)/2)} = \tan \frac{B}{2} \tan \frac{C - A}{2}$.
Substituting the expression for $\tan \frac{C - A}{2}$,we get $\tan \frac{B}{2} \left( \frac{c - a}{c + a} \cot \frac{B}{2} \right) = \frac{c - a}{c + a}$.
Substituting $c = 2a$,we get $\frac{2a - a}{2a + a} = \frac{a}{3a} = \frac{1}{3}$.
Thus,the ratio is $1:3$.

(B) Given: $a = 2$,$B = 60^\circ$,$C = 75^\circ$.
In any triangle $ABC$,the sum of angles is $180^\circ$,so $A = 180^\circ - (B + C) = 180^\circ - (60^\circ + 75^\circ) = 180^\circ - 135^\circ = 45^\circ$.
Using the Sine Rule: $\frac{b}{\sin B} = \frac{a}{\sin A}$.
Substituting the values: $\frac{b}{\sin 60^\circ} = \frac{2}{\sin 45^\circ}$.
$b = \frac{2 \times \sin 60^\circ}{\sin 45^\circ} = \frac{2 \times (\frac{\sqrt{3}}{2})}{\frac{1}{\sqrt{2}}} = \sqrt{3} \times \sqrt{2} = \sqrt{6}$.

(C) Using the Law of Sines in $\triangle ABC$:
$\frac{\sin B}{b} = \frac{\sin A}{a}$
Substitute the given values:
$\frac{\sin B}{8} = \frac{\sin 30^\circ}{6}$
Since $\sin 30^\circ = 1/2$:
$\sin B = \frac{8 \times (1/2)}{6} = \frac{4}{6} = \frac{2}{3}$
Therefore,$B = \sin^{-1}(2/3)$,which implies $x = 2/3$.

(C) Using the Law of Sines: $\frac{b}{\sin B} = \frac{c}{\sin C}$.
$\sin B = \frac{b \sin C}{c} = \frac{2 \sin 60^\circ}{\sqrt{6}} = \frac{2}{\sqrt{6}} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.
Thus,$B = 45^\circ$ (since $B + C < 180^\circ$).
Then $A = 180^\circ - (B + C) = 180^\circ - (45^\circ + 60^\circ) = 75^\circ$.
Using the Law of Sines again: $\frac{a}{\sin A} = \frac{c}{\sin C}$.
$a = \frac{c \sin A}{\sin C} = \frac{\sqrt{6} \sin 75^\circ}{\sin 60^\circ}$.
Since $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$.
$a = \frac{\sqrt{6} \times (\frac{\sqrt{6} + \sqrt{2}}{4})}{\frac{\sqrt{3}}{2}} = \frac{6 + \sqrt{12}}{4} \times \frac{2}{\sqrt{3}} = \frac{6 + 2\sqrt{3}}{2\sqrt{3}} = \frac{3}{\sqrt{3}} + 1 = \sqrt{3} + 1$.

(A) Using the Napier's analogy: $\tan \left( \frac{A - B}{2} \right) = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} = \sqrt{\frac{1 - 31/32}{1 + 31/32}} = \sqrt{\frac{1/32}{63/32}} = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.
By Napier's formula,$\frac{a - b}{a + b} \cot \frac{C}{2} = \tan \left( \frac{A - B}{2} \right)$.
Substituting the values: $\frac{5 - 4}{5 + 4} \cot \frac{C}{2} = \frac{1}{3\sqrt{7}}$ $\Rightarrow \frac{1}{9} \cot \frac{C}{2} = \frac{1}{3\sqrt{7}}$ $\Rightarrow \cot \frac{C}{2} = \frac{3}{\sqrt{7}}$.
Thus,$\tan \frac{C}{2} = \frac{\sqrt{7}}{3}$.
Using $\cos C = \frac{1 - \tan^2(C/2)}{1 + \tan^2(C/2)} = \frac{1 - 7/9}{1 + 7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.
Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 25 + 16 - 2(5)(4) \left( \frac{1}{8} \right) = 41 - 40 \left( \frac{1}{8} \right) = 41 - 5 = 36$.
Therefore,$c = \sqrt{36} = 6$.

(B) Using the Napier's Analogy: $\tan \left( \frac{C - B}{2} \right) = \frac{c - b}{c + b} \cot \left( \frac{A}{2} \right)$.
Given $A = 30^o$,$b = 2$,$c = \sqrt{3} + 1$.
$\frac{A}{2} = 15^o$.
$\cot(15^o) = \cot(45^o - 30^o) = \frac{\cot 45^o \cot 30^o + 1}{\cot 30^o - \cot 45^o} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.
Substituting the values:
$\tan \left( \frac{C - B}{2} \right) = \frac{(\sqrt{3} + 1) - 2}{(\sqrt{3} + 1) + 2} \times (2 + \sqrt{3})$
$= \frac{\sqrt{3} - 1}{\sqrt{3} + 3} \times (2 + \sqrt{3})$
$= \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} + 1)} \times (2 + \sqrt{3})$
$= \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3}(\sqrt{3} + 1)^2} \times (2 + \sqrt{3})$ (This simplifies to $\frac{1}{\sqrt{3}}$).
Thus,$\tan \left( \frac{C - B}{2} \right) = \frac{1}{\sqrt{3}} = \tan 30^o$.
Therefore,$\frac{C - B}{2} = 30^o$.

(C) Let the sides of the triangle be $a = 6 + \sqrt{12} = 6 + 2\sqrt{3}$,$b = \sqrt{48} = 4\sqrt{3}$,and $c = \sqrt{24} = 2\sqrt{6}$.
Comparing the values: $a \approx 6 + 3.46 = 9.46$,$b \approx 4 \times 1.732 = 6.928$,and $c \approx 2 \times 2.449 = 4.898$.
Since $c$ is the smallest side,the smallest angle is $C$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$a^2 = (6 + 2\sqrt{3})^2 = 36 + 12 + 24\sqrt{3} = 48 + 24\sqrt{3}$.
$b^2 = 48$.
$c^2 = 24$.
$\cos C = \frac{48 + 24\sqrt{3} + 48 - 24}{2(6 + 2\sqrt{3})(4\sqrt{3})} = \frac{72 + 24\sqrt{3}}{8\sqrt{3}(6 + 2\sqrt{3})} = \frac{24(3 + \sqrt{3})}{8\sqrt{3} \times 2(3 + \sqrt{3})} = \frac{24}{16\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Thus,$C = \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$.

(B) Given that $C = 90^\circ$,the triangle is a right-angled triangle.
Using the sine rule,we have $\frac{a}{\sin A} = \frac{c}{\sin C}$.
Substituting the given values: $a = \frac{c \sin A}{\sin C}$.
$a = \frac{7\sqrt{3} \sin 30^\circ}{\sin 90^\circ}$.
Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 90^\circ = 1$,we get:
$a = \frac{7\sqrt{3} \times (1/2)}{1} = \frac{7\sqrt{3}}{2}$.

(A) Let the angles of the triangle be $2x, 3x,$ and $7x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.
The angles are $A = 2(15^{\circ}) = 30^{\circ}$,$B = 3(15^{\circ}) = 45^{\circ}$,and $C = 7(15^{\circ}) = 105^{\circ}$.
By the Sine Rule,the ratio of the sides $a : b : c$ is $\sin A : \sin B : \sin C$.
$a : b : c = \sin 30^{\circ} : \sin 45^{\circ} : \sin 105^{\circ}$.
We know $\sin 30^{\circ} = \frac{1}{2}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,and $\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Thus,$a : b : c = \frac{1}{2} : \frac{1}{\sqrt{2}} : \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Multiplying by $2\sqrt{2}$,we get $a : b : c = \sqrt{2} : 2 : (\sqrt{3} + 1)$.

(B) Let the sides be $a = 2$,$b = \sqrt{6}$,and $c = \sqrt{3} + 1$.
We know that the smallest angle is opposite to the smallest side.
Comparing the values: $a = 2 \approx 2$,$b = \sqrt{6} \approx 2.45$,and $c = \sqrt{3} + 1 \approx 1.732 + 1 = 2.732$.
The smallest side is $a = 2$.
Using the Law of Cosines for angle $A$: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
$\cos A = \frac{(\sqrt{6})^2 + (\sqrt{3} + 1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3} + 1)} = \frac{6 + (3 + 1 + 2\sqrt{3}) - 4}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{6 + 2\sqrt{3}}{2\sqrt{6}(\sqrt{3} + 1)}$.
$\cos A = \frac{2(3 + \sqrt{3})}{2\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}(\sqrt{3} + 1)}{\sqrt{6}(\sqrt{3} + 1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.
Since $\cos A = \frac{1}{\sqrt{2}}$,we have $A = 45^o$.

(B) We know that $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\tan \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.
Substituting these into the expression:
$\frac{\tan \frac{A}{2} - \tan \frac{B}{2}}{\tan \frac{A}{2} + \tan \frac{B}{2}} = \frac{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} - \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} + \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}$
$= \frac{\frac{\sqrt{s-c}}{\sqrt{s}} \left( \frac{s-b}{\sqrt{(s-a)(s-b)}} - \frac{s-a}{\sqrt{(s-a)(s-b)}} \right)}{\frac{\sqrt{s-c}}{\sqrt{s}} \left( \frac{s-b}{\sqrt{(s-a)(s-b)}} + \frac{s-a}{\sqrt{(s-a)(s-b)}} \right)}$
$= \frac{(s-b) - (s-a)}{(s-b) + (s-a)} = \frac{a-b}{2s-a-b}$.
Since $2s = a+b+c$,we have $2s-a-b = c$.
Therefore,the expression equals $\frac{a-b}{c}$.

(B) Given: $a = \sqrt{3} + 1$,$\angle B = 30^\circ$,$\angle C = 45^\circ$.
Sum of angles in a triangle is $180^\circ$,so $\angle A = 180^\circ - (30^\circ + 45^\circ) = 105^\circ$.
The area of a triangle is given by $\Delta = \frac{a^2 \sin B \sin C}{2 \sin A}$.
Since $\sin A = \sin 105^\circ = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Substituting the values: $\Delta = \frac{(\sqrt{3} + 1)^2 \cdot \sin 30^\circ \cdot \sin 45^\circ}{2 \cdot \frac{\sqrt{3} + 1}{2\sqrt{2}}} = \frac{(\sqrt{3} + 1)^2 \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}}{2 \cdot \frac{\sqrt{3} + 1}{2\sqrt{2}}} = \frac{(\sqrt{3} + 1)^2}{4} \cdot \frac{2\sqrt{2}}{\sqrt{3} + 1} \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3} + 1}{2} \text{ cm}^2$.

(A) Let the right-angled triangle be $\Delta AOB$ with the right angle at $O$. Let $OC$ be the perpendicular drawn from the vertex $O$ to the hypotenuse $AB$,with length $OC = x$.
Let $\angle OBA = \theta$. Then $\angle OAB = 90^o - \theta$.
In $\Delta OCB$,$OC = OB \sin \theta \Rightarrow OB = \frac{x}{\sin \theta}$.
In $\Delta OCA$,$OC = OA \cos \theta \Rightarrow OA = \frac{x}{\cos \theta}$.
In $\Delta AOB$,the hypotenuse $AB = \sqrt{OA^2 + OB^2} = \sqrt{\frac{x^2}{\cos^2 \theta} + \frac{x^2}{\sin^2 \theta}} = x \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}} = \frac{x}{\sin \theta \cos \theta} = \frac{2x}{\sin 2\theta}$.
Given that the hypotenuse is $4$ times the perpendicular,$AB = 4x$.
Therefore,$\frac{2x}{\sin 2\theta} = 4x \Rightarrow \sin 2\theta = \frac{2}{4} = \frac{1}{2}$.
Since $\sin 2\theta = \sin 30^o$,we have $2\theta = 30^o$,which gives $\theta = 15^o$.
The other acute angle is $90^o - 15^o = 75^o$.

(B) Given: $\angle A = 45^\circ, \angle C = 60^\circ$.
Since the sum of angles in a triangle is $180^\circ$,$\angle B = 180^\circ - (45^\circ + 60^\circ) = 75^\circ$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,so $a = k\sin A, c = k\sin C, b = k\sin B$.
We need to evaluate $a + c\sqrt{2} = k\sin 45^\circ + k\sqrt{2}\sin 60^\circ$.
$a + c\sqrt{2} = k\left(\frac{1}{\sqrt{2}}\right) + k\sqrt{2}\left(\frac{\sqrt{3}}{2}\right) = k\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}\right) = k\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)$.
Note that $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Thus,$a + c\sqrt{2} = k(2 \sin 75^\circ) = 2k \sin B = 2b$.

(C) Let the sides of the triangle be $a = 3, b = 5,$ and $c = 7.$
Since the largest side is $c = 7,$ the largest angle is $\angle C.$
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substituting the values: $\cos C = \frac{3^2 + 5^2 - 7^2}{2 \times 3 \times 5}$
$\cos C = \frac{9 + 25 - 49}{30} = \frac{-15}{30} = -\frac{1}{2}$
Since $\cos C = -\frac{1}{2},$ we have $\angle C = \arccos(-1/2) = \frac{2\pi}{3}.$

(A) Applying the sine rule,$\frac{\sin B}{b} = \frac{\sin A}{a}$.
Substituting the given values,$\frac{\sin B}{8} = \frac{5/13}{3}$.
This simplifies to $\sin B = \frac{40}{39}$.
Since $\frac{40}{39} \approx 1.025 > 1$,and the value of $\sin B$ cannot exceed $1$,no such angle $B$ exists.
Therefore,no triangle $ABC$ can be formed with the given conditions.

(B) We know that in a $\Delta ABC,$ $A + B + C = \pi,$ so $A + C = \pi - B.$
Substituting this into the expression: $\frac{A - B + C}{2} = \frac{(A + C) - B}{2} = \frac{(\pi - B) - B}{2} = \frac{\pi - 2B}{2} = \frac{\pi}{2} - B.$
Thus,$2ac \sin \left( \frac{\pi}{2} - B \right) = 2ac \cos B.$
Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}.$
Therefore,$2ac \cos B = 2ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = a^2 + c^2 - b^2.$

(D) Given a right-angled triangle $ABC$ with the right angle at $C$.
Let $a, b,$ and $c$ be the lengths of sides $BC, CA,$ and $AB$ respectively.
From the Pythagoras theorem,we have $c^2 = a^2 + b^2$.
In the right-angled triangle $ABC$:
$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{a}{b}$
$\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{BC} = \frac{b}{a}$
Therefore,$\tan A + \tan B = \frac{a}{b} + \frac{b}{a}$
$= \frac{a^2 + b^2}{ab}$
Since $a^2 + b^2 = c^2$,we get:
$= \frac{c^2}{ab}$

(C) Given $A:B:C = 3:5:4.$ Let $A = 3x, B = 5x, C = 4x.$
Since $A + B + C = 180^{\circ},$ we have $12x = 180^{\circ},$ so $x = 15^{\circ}.$
Thus,$A = 45^{\circ}, B = 75^{\circ}, C = 60^{\circ}.$
Using the Sine Rule,$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 75^{\circ}} = \frac{c}{\sin 60^{\circ}} = K.$
Then $a = K \sin 45^{\circ} = \frac{K}{\sqrt{2}},$
$b = K \sin 75^{\circ} = K \sin(45^{\circ} + 30^{\circ}) = K(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}) = \frac{K(\sqrt{3} + 1)}{2\sqrt{2}},$
$c = K \sin 60^{\circ} = \frac{K\sqrt{3}}{2}.$
Now,$a + b + c\sqrt{2} = \frac{K}{\sqrt{2}} + \frac{K(\sqrt{3} + 1)}{2\sqrt{2}} + \frac{K\sqrt{3}}{2} \cdot \sqrt{2} = \frac{K}{\sqrt{2}} + \frac{K\sqrt{3} + K}{2\sqrt{2}} + \frac{K\sqrt{3}}{\sqrt{2}} = \frac{2K + K\sqrt{3} + K + 2K\sqrt{3}}{2\sqrt{2}} = \frac{3K + 3K\sqrt{3}}{2\sqrt{2}} = \frac{3K(1 + \sqrt{3})}{2\sqrt{2}}.$
Since $3b = 3 \cdot \frac{K(\sqrt{3} + 1)}{2\sqrt{2}},$ we see that $a + b + c\sqrt{2} = 3b.$

(B) Given expression: $\frac{\cos C + \cos A}{c + a} + \frac{\cos B}{b}$
Taking the common denominator $b(c + a)$,we get:
$= \frac{b(\cos C + \cos A) + \cos B(c + a)}{b(c + a)}$
$= \frac{b \cos C + b \cos A + c \cos B + a \cos B}{b(c + a)}$
Using the projection formula,we know that $b \cos C + c \cos B = a$ and $b \cos A + a \cos B = c$.
Substituting these values:
$= \frac{(b \cos C + c \cos B) + (b \cos A + a \cos B)}{b(c + a)}$
$= \frac{a + c}{b(c + a)}$
$= \frac{1}{b}$.

(A) Let the angles of the triangle be $x$,$3x$,and $5x$.
Since the sum of the angles in a triangle is $180^{\circ}$ or $\pi$ radians,we have:
$x + 3x + 5x = \pi$
$9x = \pi$
$x = \frac{\pi}{9}$
The greatest angle is $5x = 5 \times \frac{\pi}{9} = \frac{5\pi}{9}$.

(D) In $\triangle ABC$,by the Law of Cosines in $\triangle ABC$ at angle $B$:
$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC} = \frac{2^2 + 4^2 - 3^2}{2 \times 2 \times 4} = \frac{4 + 16 - 9}{16} = \frac{11}{16}$.
Since $D$ is the midpoint of $BC$,$BD = 2$.
Applying the Law of Cosines in $\triangle ABD$:
$AD^2 = AB^2 + BD^2 - 2 \times AB \times BD \times \cos B$
$AD^2 = 2^2 + 2^2 - 2 \times 2 \times 2 \times \frac{11}{16}$
$AD^2 = 4 + 4 - 8 \times \frac{11}{16}$
$AD^2 = 8 - \frac{11}{2} = 8 - 5.5 = 2.5$.
Thus,$AD^2 = 2.5$.

(B) Given that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$
This implies $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2}$.
Using the identity $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$,we have $\cot \frac{A}{2} = \frac{s-a}{r}$,where $r$ is the inradius and $s$ is the semi-perimeter.
Substituting this into the $A.P.$ condition: $2 \frac{s-b}{r} = \frac{s-a}{r} + \frac{s-c}{r}$.
$2(s-b) = s-a + s-c \implies 2s - 2b = 2s - (a+c) \implies 2b = a+c$.
This means $a, b, c$ are in $A.P.$
For an equilateral triangle where $A=B=C=60^{\circ}$,$\cot 30^{\circ} = \sqrt{3}$.
Then $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{3} \times \sqrt{3} = 3$. Thus,option $(b)$ is correct.

(A) The smallest angle is opposite to the smallest side. Comparing the sides: $a = 7$,$b = 4\sqrt{3} \approx 6.928$,and $c = \sqrt{13} \approx 3.605$.
Since $c < b < a$,the smallest angle is $C$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$\cos C = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 \times 7 \times 4\sqrt{3}}$.
$\cos C = \frac{49 + 48 - 13}{56\sqrt{3}} = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
Since $\cos C = \frac{\sqrt{3}}{2}$,we have $C = 30^o$.

(B) Let $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = \lambda.$
Then $b + c = 11\lambda$ $(i), c + a = 12\lambda$ $(ii),$ and $a + b = 13\lambda$ $(iii).$
Adding $(i), (ii),$ and $(iii),$ we get $2(a + b + c) = 36\lambda,$ so $a + b + c = 18\lambda$ $(iv).$
Subtracting $(i)$ from $(iv),$ $a = 7\lambda.$
Subtracting $(ii)$ from $(iv),$ $b = 6\lambda.$
Subtracting $(iii)$ from $(iv),$ $c = 5\lambda.$
Using the cosine rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}.$
Substituting the values,$\cos C = \frac{(7\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2(7\lambda)(6\lambda)} = \frac{49\lambda^2 + 36\lambda^2 - 25\lambda^2}{84\lambda^2} = \frac{60\lambda^2}{84\lambda^2} = \frac{5}{7}.$

(B) Given: $b = 20, c = 21, \sin A = 3/5$.
Since $\sin^2 A + \cos^2 A = 1$,we have $\cos A = \pm \sqrt{1 - (3/5)^2} = \pm 4/5$.
Assuming $A$ is an acute angle,$\cos A = 4/5$.
Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$.
$a^2 = 20^2 + 21^2 - 2(20)(21)(4/5)$.
$a^2 = 400 + 441 - 840(4/5) = 841 - 672 = 169$.
Therefore,$a = \sqrt{169} = 13$.

(B) Let the side length of the equilateral triangle $ADC$ be $x$. Thus,$DC = AD = AC = x$.
Since $D$ is the midpoint of $BC$,$BD = DC = x$,so $BC = a = 2x$.
In $\triangle ADC$,all angles are $60^\circ$. Since $\angle ADC = 60^\circ$,the adjacent angle $\angle ADB = 180^\circ - 60^\circ = 120^\circ$.
In $\triangle ABD$,by the Law of Cosines:
$AB^2 = BD^2 + AD^2 - 2(BD)(AD) \cos(120^\circ)$
$c^2 = x^2 + x^2 - 2(x)(x)(-1/2)$
$c^2 = 2x^2 + x^2 = 3x^2$.
We have $a = 2x$,$b = AC = x$,and $c^2 = 3x^2$.
Therefore,$a^2 : b^2 : c^2 = (2x)^2 : x^2 : 3x^2 = 4x^2 : x^2 : 3x^2 = 4:1:3$.

(D) Given the ratio of sides $a:b:c = 1:\sqrt{3}:2$.
Let $a = \lambda$,$b = \sqrt{3}\lambda$,and $c = 2\lambda$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3\lambda^2 + 4\lambda^2 - \lambda^2}{2(\sqrt{3}\lambda)(2\lambda)} = \frac{6\lambda^2}{4\sqrt{3}\lambda^2} = \frac{\sqrt{3}}{2}$.
Thus,$A = 30^\circ$.
Similarly,$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{\lambda^2 + 4\lambda^2 - 3\lambda^2}{2(\lambda)(2\lambda)} = \frac{2\lambda^2}{4\lambda^2} = \frac{1}{2}$.
Thus,$B = 60^\circ$.
Finally,$C = 180^\circ - (30^\circ + 60^\circ) = 90^\circ$.
The ratio $A:B:C = 30^\circ:60^\circ:90^\circ = 1:2:3$.

(D) Given: $b = \sqrt{3}$,$c = 1$,and $\angle A = 30^\circ$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\cos 30^\circ = \frac{(\sqrt{3})^2 + 1^2 - a^2}{2(\sqrt{3})(1)}$.
$\frac{\sqrt{3}}{2} = \frac{3 + 1 - a^2}{2\sqrt{3}}$.
$3 = 4 - a^2$ $\Rightarrow a^2 = 1$ $\Rightarrow a = 1$.
Since $b = \sqrt{3} \approx 1.732$ and $a = c = 1$,the side $b$ is the largest side.
The largest angle is opposite the largest side,which is $\angle B$.
Using the Law of Cosines for $\angle B$: $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2(1)(1)} = \frac{1 + 1 - 3}{2} = -\frac{1}{2}$.
Since $\cos B = -\frac{1}{2}$,we have $B = 120^\circ$.

(C) Let the sides be $a = \alpha - \beta$,$b = \alpha + \beta$,and $c = \sqrt{3\alpha^2 + \beta^2}$.
Since $\alpha > \beta > 0$,we observe that $c^2 = 3\alpha^2 + \beta^2$ is the largest square,as $a^2 + b^2 = 2\alpha^2 + 2\beta^2 < 3\alpha^2 + \beta^2$ (since $\alpha^2 > \beta^2$).
Thus,$c$ is the longest side,and the angle $C$ opposite to $c$ is the largest angle.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{(\alpha - \beta)^2 + (\alpha + \beta)^2 - (3\alpha^2 + \beta^2)}{2(\alpha - \beta)(\alpha + \beta)}$.
$\cos C = \frac{(\alpha^2 - 2\alpha\beta + \beta^2) + (\alpha^2 + 2\alpha\beta + \beta^2) - 3\alpha^2 - \beta^2}{2(\alpha^2 - \beta^2)}$.
$\cos C = \frac{2\alpha^2 + 2\beta^2 - 3\alpha^2 - \beta^2}{2(\alpha^2 - \beta^2)} = \frac{\beta^2 - \alpha^2}{2(\alpha^2 - \beta^2)}$.
$\cos C = \frac{-(\alpha^2 - \beta^2)}{2(\alpha^2 - \beta^2)} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,the angle $C = \frac{2\pi}{3}$.

(B) We know the half-angle formulas for a triangle:
$\cos^2 \frac{A}{2} = \frac{s(s - a)}{bc}$
$\sin^2 \frac{A}{2} = \frac{(s - b)(s - c)}{bc}$
Substituting these into the given expression:
$\frac{s(s - a)}{bc} - \frac{(s - b)(s - c)}{bc} = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}$
Using the identity $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$,we get:
$\cos^2 \frac{A}{2} - \sin^2 \frac{A}{2} = \cos \left( 2 \times \frac{A}{2} \right) = \cos A$

(C) Given: $a = 2$,$b = 4$,and $\angle C = 60^\circ$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$\cos 60^\circ = \frac{2^2 + 4^2 - c^2}{2(2)(4)} \Rightarrow \frac{1}{2} = \frac{4 + 16 - c^2}{16}$.
$8 = 20 - c^2$ $\Rightarrow c^2 = 12$ $\Rightarrow c = 2\sqrt{3}$.
Using the Sine Rule: $\frac{a}{\sin A} = \frac{c}{\sin C}$ $\Rightarrow \sin A = \frac{a \sin C}{c} = \frac{2 \sin 60^\circ}{2\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$.
Thus,$\angle A = 30^\circ$.
Since $\angle A + \angle B + \angle C = 180^\circ$,we have $30^\circ + \angle B + 60^\circ = 180^\circ$.
Therefore,$\angle B = 90^\circ$.

(B) The area of a triangle $ABC$ with sides $a, b, c$ and corresponding angles $A, B, C$ is given by the formula:
$\text{Area} = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C$.
Thus,the correct expression among the given options is $\frac{1}{2}bc \sin A$.

(C) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a : b : c = \sin A : \sin B : \sin C = 1 : 2 : 3$.
Let $a = x$,$b = 2x$,and $c = 3x$.
Given $b = 4 \, \text{cm}$,we have $2x = 4$,which implies $x = 2 \, \text{cm}$.
Therefore,$a = 2 \, \text{cm}$,$b = 4 \, \text{cm}$,and $c = 6 \, \text{cm}$.
The perimeter of the triangle is $a + b + c = 2 + 4 + 6 = 12 \, \text{cm}$.

(C) Let the sides of the triangle be $5x, 12x, 13x$.
Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.
The area of a right-angled triangle is given by $\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (5x) \times (12x)$.
Given $\Delta = 270 \text{ cm}^2$,we have $30x^2 = 270$.
$x^2 = 9$,which implies $x = 3$.
Therefore,the sides are $5(3), 12(3), 13(3)$,which are $15 \text{ cm}, 36 \text{ cm}, 39 \text{ cm}$.

(B) Given $\cos A = \frac{\sin B}{2\sin C}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation:
$\cos A = \frac{b/2R}{2(c/2R)} = \frac{b}{2c}$.
Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Equating the two expressions for $\cos A$:
$\frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{2c}$.
Multiplying both sides by $2bc$:
$b^2 + c^2 - a^2 = b^2$.
$c^2 - a^2 = 0$ $\Rightarrow c^2 = a^2$ $\Rightarrow c = a$.
Since two sides are equal,the triangle is isosceles.

(B) The area of a triangle with two sides $b$ and $c$ and included angle $A$ is given by $\Delta = \frac{1}{2}bc \sin A$.
Given $\Delta = 9 \, cm^2$ and $b = c = 6 \, cm$.
Substituting these values,we get $9 = \frac{1}{2} \times 6 \times 6 \times \sin A$.
$9 = 18 \sin A$.
$\sin A = \frac{9}{18} = \frac{1}{2}$.
Since $\sin A = \frac{1}{2}$,the angle $A = 30^\circ$.

(A) Let the sides be $a = 6$,$b = 10$,and $c = 14$.
To determine the type of triangle,we use the Law of Cosines to find the angle $\theta$ opposite to the longest side $c = 14$.
$\cos \theta = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos \theta = \frac{6^2 + 10^2 - 14^2}{2 \times 6 \times 10}$
$\cos \theta = \frac{36 + 100 - 196}{120}$
$\cos \theta = \frac{136 - 196}{120} = \frac{-60}{120} = -\frac{1}{2}$
Since $\cos \theta = -\frac{1}{2}$,we have $\theta = 120^\circ$.
Since one angle is greater than $90^\circ$,the triangle is an obtuse-angled triangle.

(B) Using the Sine Rule,we have $a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these into the given equation $a \cos B = b \cos A$:
$(2R \sin A) \cos B = (2R \sin B) \cos A$
$\sin A \cos B = \sin B \cos A$
$\sin A \cos B - \cos A \sin B = 0$
$\sin(A - B) = 0$
Since $A$ and $B$ are angles of a triangle,$A - B = 0$,which implies $A = B$.
Therefore,the triangle is an isosceles triangle.