Properties of Triangles

$ \begin{aligned} & \text { } \begin{aligned} \sin A & =\frac{3 \sqrt{3}}{14}=\frac{\sqrt{27}}{14} \\ \cos A & =\sqrt{1-\sin ^2 A}\,\,\,\,\,\,\, \left[\because A<90^{\circ}\right]\\ & =\sqrt{1-\frac{27}{196}}=\frac{13}{14} \end{aligned} \end{aligned} $

By cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$

$ \begin{array}{ll} \Rightarrow & \frac{13}{14}=\frac{25+c^2-9}{2 \times 5 \times c} \\ \Rightarrow & 7 c^2-65 c+112=0 \\ \Rightarrow & 7 c^2-49 c-16 c+112=0 \\ \Rightarrow & \quad(c-7)(7 c-16)=0 \\ \Rightarrow & \quad c=7,16 / 7 \\ \therefore & \quad c=7 \end{array} $

[ $\because c$ is obtuse; $\therefore c$ is greatest]

$ \begin{aligned} &\begin{aligned} & (b \sin C+c \sin B)(b \cos C+c \cos B) \\ & =(b \sin C+c \sin B)(a) \end{aligned}\\ &\text { [By projection formula] }\\ &\begin{aligned} & =a b \sin C+a c \sin B \\ & =2 \Delta+2 \Delta \\ & {\left[\because \Delta=\frac{1}{2} a b \sin C \text { and } \Delta=\frac{1}{2} a c \sin B\right]} \\ & =4 \Delta \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} \\ & \because \quad A=\pi r^2 \Rightarrow r=\sqrt{A / \pi} \\ & \therefore \quad r_1=\sqrt{A_1 / \pi} \\ & \quad r_2=\sqrt{\frac{A_2}{\pi}} \text { and } r_3=\sqrt{\frac{A_3}{\pi}} \\ & \Rightarrow \quad \frac{\sqrt{\pi}}{\sqrt{A}}=\frac{\sqrt{\pi}}{\sqrt{A_1}}+\frac{\sqrt{\pi}}{\sqrt{A_2}}+\frac{\sqrt{\pi}}{\sqrt{A_3}} \\ & \Rightarrow \quad \frac{1}{\sqrt{A}}=\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}} \\ & \Rightarrow \quad \frac{1}{\sqrt{A}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12} \Rightarrow A=\frac{144}{169} \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { In } \triangle A B C, \\ & \begin{aligned} (b+c)^2 \sin ^2 \frac{A}{2}+(b-c)^2 \cos ^2 \frac{A}{2} & \\ = & K(1-\cos 2 A) \\ \Rightarrow(b+c)^2 \sin ^2 \frac{A}{2}+(b-c)^2( & \left.1-\sin ^2 \frac{A}{2}\right) \\ = & K\left(2 \sin ^2 A\right) \\ \Rightarrow(b-c)^2+\left\{(b+c)^2-(b-c)^2\right\} & \sin ^2 \frac{A}{2} \\ = & 2 K \sin ^2 A \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { On putting }\\ &\begin{aligned} & \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}, \sin A=\frac{a}{2 R} \\ & \text { So, }(b-c)^2+4 b c\left(\frac{(s-b)(s-c)}{b c}\right) \\ & =2 K\left(\frac{a^2}{4 R^2}\right) \\ & \Rightarrow(b-c)^2+4\left(\frac{a+b+c}{2}-b\right) \\ & \left(\frac{a+b+c}{2}-c\right)=\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+(a+c-b)(a+b-c)=\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+[a-(b-c)][a+(b-c)] \\ & =\frac{K a^2}{2 R^2} \\ & \Rightarrow(b-c)^2+\left[a^2-(b-c)^2\right]=\frac{K a^2}{2 R^2} \\ & \Rightarrow a^2=\frac{K a^2}{2 R^2} \Rightarrow K=2 R^2 \end{aligned} \end{aligned} $

In $\triangle A B C, b=7, c=4 \sqrt{3}$ and $A=\frac{\pi}{6}$

$ \begin{aligned} & \because \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow \cos \frac{\pi}{6}=\frac{49+48-a^2}{2 \times 7 \times 4 \sqrt{3}} \\ & \Rightarrow \frac{\sqrt{3}}{2}=\frac{97-a^2}{56 \sqrt{3}} \\ & \Rightarrow 84=97-a^2 \\ & a^2=13 \text { and } a=\sqrt{13} \end{aligned} $

Using sine rule,

$ \begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ & \frac{\sin \frac{\pi}{6}}{\sqrt{13}}=\frac{\sin B}{7}=\frac{\sin C}{4 \sqrt{3}} \\ & \sin B=\frac{7}{2 \sqrt{13}}, \sin C=\frac{2 \sqrt{3}}{\sqrt{13}} \end{aligned} $

$ \begin{aligned} \therefore a \sin B \sin C & =\sqrt{13} \times \frac{7}{2 \sqrt{13}} \times \frac{2 \sqrt{3}}{\sqrt{13}} \\ & =\frac{7 \sqrt{3}}{\sqrt{13}} \end{aligned} $

$ \begin{aligned} &\text { In } \triangle A B C \text {, }\\ &\begin{aligned} & r_1=\frac{\Delta}{s-a^{\prime}}, r_2=\frac{\Delta}{s-b} \text { and } r_3=\frac{\Delta}{s-c} \\ \because & 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{2 s-b-c}{(s-b)(s-c)} \times \frac{s(s-a)}{s(s-a)}\right) \\ = & \Delta\left(\frac{a s(s-a)}{\Delta^2}\right) \\ & r_2+r_3=\frac{a s(s-a)}{\Delta}=\frac{a}{\tan A / 2}=a \\ \quad & \quad\left[\because A=\frac{\pi}{2}\right] \end{aligned} \end{aligned} $

Given that,

$ \begin{equation*} \frac{a^3+b^3+c^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Using sine rule for a triangle $A B C$

$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R $

{where $2 R \rightarrow$ circumradius }

$\therefore a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$

Substituting the values of $a, b, c$ in Eq

(i), we get

$\frac{(2 R \sin A)^3+(2 R \sin B)^3+(2 R \sin C)^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$

or $\frac{(2 R)^3\left(\sin ^3 A+\sin ^3 B+\sin ^3 C\right)}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$

or $(2 R)^3=8 \quad$ or $\quad 2 R=2$

For a triangle $A B C$, it is given that

$ c=9, s=10 $

$ \begin{aligned} \text { and } \quad \Delta & =10 \sqrt{2} \\ \therefore \tan \frac{A-B}{2} & =\frac{a-b}{a+b} \cot \frac{c}{2}=\frac{a-b}{a+b} \frac{s(s-c)}{\Delta} \\ & =\frac{a-b}{a+b} \times \frac{10 \times 1}{10 \sqrt{2}}=\frac{1}{\sqrt{2}} \frac{a-b}{a+b} \end{aligned} $

So, $b\left[1+\sqrt{2} \tan \frac{A-B}{2}\right]$

$ =b\left[1+\left(\sqrt{2} \times \frac{1}{\sqrt{2}} \frac{a-b}{a+b}\right)\right] $

$ \begin{aligned} & =b\left[\frac{a+b+a-b}{a+b}\right] \\ & =b\left[\frac{2 a}{a+b}\right]=a\left[\frac{2 b}{a+b}\right]=a\left[\frac{(a+b)-(a-b)}{a+b}\right] \\ & =a\left[1-\frac{a-b}{a+b}\right]=a\left[1-\sqrt{2} \tan \frac{A-B}{2}\right] \end{aligned} $

Let a $\triangle A B C$ of sides $a, b, c$, having area △

Area $=\Delta$

$ \frac{1}{2} b c \sin A=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

and $\quad \frac{1}{2} a c \sin B=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $\quad \frac{1}{2} a b \sin c=\Delta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using cosine rule

$ a^2=b^2+c^2-2 b c \cos A $

and    $ b^2=a^2+c^2-2 a c \cos B $

and     $ c^2=a^2+b^2-2 a b \cos C $

On adding, we get

$ \begin{align*} & \quad a^2+b^2+c^2=2 a^2+2 b^2+2 c^2 \\ & \quad-2 a b \cos C-2 a c \cos B-2 b c \cos A \\ & \text { or } \quad a^2+b^2+c^2=2 \\ & (a b \cos C+a c \cos B+b c \cos A) \quad \text { } \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Now, from Eq. (i)

$ b c=\frac{2 \Delta}{\sin A} $

Eq. (ii), $\quad a c=\frac{2 \Delta}{\sin B}$

Eq. (iii), $ a b=\frac{2 \Delta}{\sin C} $

Putting these values in Eq. (iv), we get

$ \begin{aligned} & a^2+b^2+c^2=2 \\ & \left(\frac{2 \Delta}{\sin C} \cos C+\frac{2 \Delta}{\sin B} \cos B+\frac{2 \Delta}{\sin A} \cos A\right) \\ & a^2+b^2+c^2=4 \Delta(\cot C+\cot B+\cot A) \\ & \text { or } \cot C+\cot B+\cot A=\frac{a^2+b^2+c^2}{4 \Delta} \\ & \text { or } \cot A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4 \Delta} \end{aligned} $

Let three sides are $x-1, x, x+1$.

Given, largest angle is twice the smallest angle

$ \begin{align*} \therefore \cos \theta & =\frac{(x+1)^2+(x)^2-(x-1)^2}{2 x(x+1)} \\ & =\frac{x^2+2 x+1+x^2-x^2+2 x-1}{2 x(x+1)} \\ & =\frac{x^2+4 x}{2 x(x+1)}=\frac{x+4}{2(x+1)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

By Sine rule,

$ \begin{align*} & & \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)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eqs. (i) and (ii),

$ \begin{aligned} & & \frac{x+1}{2(x-1)} & =\frac{x+4}{2(x+1)} \\ \Rightarrow & & x & =5 \end{aligned} $

∴ Sides are 4, 5, 6

$ \begin{aligned} \Delta & =\frac{1}{2} \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} \\ & =\frac{1}{4} \sqrt{15 \times 7 \times 5 \times 3}=\frac{15}{4} \sqrt{7} \end{aligned} $

In $\triangle A B C, A D$ and $B E$ are medians of $\triangle A B C$

$ \begin{array}{rlrl} A D & =\frac{7}{2} \Rightarrow \angle D A B=\frac{\pi}{8} \\ \angle A B E & =\frac{\pi}{4} \\ \Rightarrow & A O =\frac{2}{3} A D=\frac{2}{3} \times \frac{7}{2}=\frac{7}{3} \end{array} $

$ \angle A O B=\pi-\left(\frac{\pi}{8}+\frac{\pi}{4}\right)=\left(\pi-\frac{3 \pi}{8}\right) $

By Sine rule, $\frac{A O}{\sin \frac{\pi}{4}}=\frac{B O}{\sin \frac{\pi}{8}}$

$ \begin{aligned} B O=\frac{\frac{7}{3} \times \sin \frac{\pi}{8}}{\sin \frac{\pi}{4}} & =\frac{7}{3} \times \frac{\sin \frac{\pi}{8}}{2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}} \\ & =\frac{7}{6 \cos \frac{\pi}{8}} \end{aligned} $

Area of $\triangle A O B=\frac{1}{2} A O \times B O \times \sin \angle A O B$

$ \begin{aligned} & =\frac{1}{2} \times \frac{7}{3} \times \frac{7}{6} \times \frac{\sin \left(\pi-\frac{3 \pi}{8}\right)}{\cos \frac{\pi}{8}} \\ & =\frac{7}{36} \times \frac{\sin 3 \frac{\pi}{8}}{\cos \frac{\pi}{8}}=\frac{7}{36} \\ & \quad\left[\because \sin \frac{3 \pi}{8}=\cos \frac{3 \pi}{8}\right] \end{aligned} $

Area of $\triangle A B C=3 \times \frac{7}{36}=\frac{7}{12}$

Given, sides of $\triangle A B C$ is $5 k, 5 k$ and $6 k$ respectively,

$ \begin{aligned} &A B C \text { is an isosceles triangle, }\\ &\begin{aligned} & \Rightarrow \quad A D=\sqrt{25 k^2-9 k^2}=4 k \\ & \Rightarrow \quad \tan \theta=\frac{3}{4} \\ &\,\,\,\, \quad \angle B A C=2 \theta \\ & \Rightarrow \quad \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\ & \Rightarrow \quad \tan 2 \theta=\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7} \\ & \Rightarrow \quad 2 \theta=\tan ^{-1}\left(\frac{24}{7}\right) \\ & \therefore \text { Largest angle is } \tan ^{-1}\left(\frac{24}{7}\right) . \end{aligned} \end{aligned} $

In $\triangle A B C$,

$ \left(b^2-c^2\right) \cot A+\left(c^2-a^2\right) \cot B $

We know that,

$ \begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R \\ & \begin{array}{r} \because 4 R^2\left[\left(\sin ^2 B-\sin ^2 C\right) \cot A\right. \\ \left.\quad+\left(\sin ^2 C-\sin ^2 A\right) \cot B\right] \end{array} \\ & 4 R^2[\sin (B+C) \sin (B-C)] \\ & \left.\frac{\cos A}{\sin A}+\sin (C+A) \sin (C-A) \frac{\cos B}{\sin B}\right] \\ & 4 R^2\left[\frac{\sin A \sin (B-C) \cos A}{\sin A}\right. \left.+\frac{\sin B \sin (C-A) \cos B}{\sin B}\right] \\ & 4 R^2[-\sin (B-C) \cos (B+C)-\sin (C-A) \cos (C+A)] \\ & 2 R^2[-\sin 2 B+\sin 2 C+\sin 2 A-\sin 2 C] \\ & 2 R^2[\sin 2 A-\sin 2 B] \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \frac{\Delta^2}{a^2+b^2+c^2}\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}\right) \\ & =\frac{\Delta^2}{a^2+b^2+c^2} \\ & {\left[\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2}+\frac{s^2}{\Delta^2}\right]} \\ & =\frac{1}{a^2+b^2+c^2} \\ & {\left[(s-a)^2+(s-b)^2+(s-c)^2+s^2\right]} \\ & =\frac{1}{a^2+b^2+c^2} \end{aligned} \end{aligned} $

$ \begin{aligned} & {\left[\left(\frac{a+b+c}{2}-a\right)^2+\left(\frac{a+b+c}{2}-b\right)^2\right.} \\ & \left.\quad+\left(\frac{a+b+c}{2}-c\right)^2+\left(\frac{a+b+c}{2}\right)^2\right] \\ & =\frac{1}{a^2+b^2+c^2} \\ & {\left[\left(\frac{b+c-a}{2}\right)^2+\left(\frac{a+c-b}{2}\right)^2\right.} \\ & \left.\quad+\left(\frac{a+b-c}{2}\right)^2+\left(\frac{a+b+c}{2}\right)^2\right] \end{aligned} $

$ =\frac{1}{a^2+b^2+c^2} $

$ =\frac{1}{a^2+b^2+c^2}\left[\frac{4\left(a^2+b^2+c^2\right)}{4}\right]=1 $

We have,

$ \begin{aligned} & R: r_1: r: 5: 12: 2 \\ & \therefore R=5 k, r_1=12 k, \\ & r=2 k \end{aligned} $

If $\triangle A B C$ is right angle of $A$

and $a=10 k, b=6 k$,

$ c=8 k $

Then $R=5 k, r_1=12 k$

Possible

$ \begin{array}{ll} \therefore & a^2=b^2+c^2 \\ \therefore & r+r_3+r_2-r_1 \end{array} $

$ \begin{aligned} & \frac{\Delta}{s}+\frac{\Delta}{s-c}+\frac{\Delta}{s-b}+\frac{\Delta}{s-a} \\ & \left(\frac{\Delta}{s}-\frac{\Delta}{s-a}\right)+\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-b}\right) \\ & \Rightarrow \Delta\left[\frac{s-a-s}{s(s-a)}+\frac{s-b+s-c}{(s-c)(s-b)}\right) \\ & \Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{2 s-b-c}{(s-c)(s-b)}\right] \\ & \Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{a}{(s-c)(s-b)}\right] \\ & \Rightarrow \Delta a\left[\frac{-(s-c)(s-b)+s(s-a)}{s(s-a)(s-b)(s-c)}\right] \\ & \Rightarrow \Delta a \frac{\left(s^2-a s-s^2+(b+c) s-b c\right)}{\Delta^2} \\ & \Rightarrow \frac{a^2-\left(b^2+c^2\right)}{\Delta^2}=0=\cos A \\ & \therefore \cos A \end{aligned} $

In $\triangle A B C$, we have

$ \begin{aligned} & \frac{A B}{\sin C}=\frac{B C}{\sin A}=\frac{A C}{\sin B} \\ \Rightarrow \quad & \frac{A B}{\sin \left(180^{\circ}-4 \theta\right)}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \\ \Rightarrow \quad & \frac{A B}{\sin 4 \theta}=\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta} \end{aligned} $

Now, $\frac{16}{\sin 3 \theta}=\frac{9}{\sin \theta}$

$ \begin{aligned} & \Rightarrow \quad 16 \sin \theta=9 \sin 3 \theta \\ & \Rightarrow \quad 16 \sin \theta=9\left(3 \sin \theta-4 \sin ^3 \theta\right) \\ & \Rightarrow \quad 16=27-36 \sin ^2 \theta \Rightarrow \\ & 36 \sin ^2 \theta=11 \\ & \Rightarrow \quad \sin \theta=\frac{\sqrt{11}}{6} \Rightarrow \cos \theta=\frac{5}{6} \end{aligned} $

Now, $\sin 4 \theta=4 \sin \theta \cos \theta\left(1-2 \sin ^2 \theta\right)$

$ =4 \times \frac{\sqrt{11}}{6} \times \frac{5}{6}\left(1-2 \times \frac{11}{36}\right)=\frac{35 \sqrt{11}}{162} $

Again, we have

$ \begin{aligned} \frac{A B}{\sin 4 \theta} & =\frac{9}{\sin \theta} \\ \Rightarrow \quad A B & =\frac{9 \sin 4 \theta}{\sin \theta}=9 \times \frac{35 \sqrt{11}}{162} \times \frac{6}{\sqrt{11}} \\ & =\frac{35}{3} \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} r_1+r_2=4 R\left(\sin \frac{A}{2} \cos \frac{B}{2}\right. & \cos \frac{C}{2} \left.+\sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =4 R \cos \frac{C}{2}\left[\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}\right] \\ & =4 R \cos \frac{C}{2} \sin \left(\frac{A+B}{2}\right) \\ & =4 R \cos \frac{C}{2} \sin \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =4 R \cos \frac{C}{2} \cdot \cos \frac{C}{2} \\ & =2 R\left(2 \cos ^2 \frac{C}{2}\right)=2 R(1+\cos C) \\ & \therefore \quad \frac{1+\cos C}{r_1+r_2}=\frac{1}{2 R} \end{aligned}\\ &\text { Similarly, }\\ &\begin{aligned} & \frac{1+\cos A}{r_2+r_3}=\frac{1}{2 R}, \frac{1+\cos B}{r_1+r_3}=\frac{1}{2 R} \\ \therefore \quad & \frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3} +\frac{1+\cos B}{r_1+r_3}=\frac{3}{2 R} \end{aligned} \end{aligned} $

We have,

$ \begin{array}{r} \cos A \cos B+\sin A \sin B \sin C=1 \\ \Rightarrow \quad \cos A \cos B+\sin A \sin B-\sin A \sin B \\ +\sin A \sin B \sin C=1 \\ \Rightarrow \cos (A-B)-\sin A \sin B(1-\sin C)=1 \\ \Rightarrow 1-\cos (A-B)+\sin A \sin B(1-\sin C) =0 \\ \Rightarrow 2 \sin ^2\left(\frac{A-B}{2}\right)+\sin A \sin B(1-\sin C)=0 \end{array} $

⇒ Now, $0 \leq \sin A, \sin B, \sin C \leq 1$

$ \therefore \quad \sin \left(\frac{A-B}{2}\right)=0 $

and $\sin A \sin B(1-\sin C)=0$

$ \Rightarrow A-B=0 \text { and } \sin C=1 $

$ \begin{array}{lc} & {[\because \sin A, \sin B \neq 0]} \\ \Rightarrow & A=B \text { and } C=90^{\circ} \\ \therefore & A=B=45^{\circ}, C=90^{\circ} \\ \text { Now, } & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ \Rightarrow & \frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 45^{\circ}}=\frac{c}{\sin 90^{\circ}} \\ \Rightarrow & a: b: c=\frac{1}{\sqrt{2}}: \frac{1}{\sqrt{2}}: 1=1: 1: \sqrt{2} \end{array} $