$$\eqalign{ & 2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2} = 2{\sin ^2}\frac{C}{2} \cr & \Rightarrow \,\,\cos \frac{{A - B}}{2} = 2\cos \frac{{A + B}}{2}\,\,\,\,\left( {\because \,\,\cos \frac{{A + B}}{2} \ne 0} \right) \cr & {\text{or, }}\cos \frac{A}{2} \cdot \cos \frac{B}{2} + \sin \frac{A}{2} \cdot \sin \frac{B}{2} = 2\,\,\left( {\cos \frac{A}{2} \cdot \cos \frac{B}{2} - \sin \frac{A}{2} \cdot \sin \frac{B}{2}} \right) \cr & {\text{or, }}3\sin \frac{A}{2} \cdot \sin \frac{B}{2} = \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cr & \Rightarrow \,\,\tan \frac{A}{2} \cdot \tan \frac{B}{2} = \frac{1}{3} \cr & {\text{or, }}\sqrt {\frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} \cdot \sqrt {\frac{{\left( {s - c} \right)\left( {s - a} \right)}}{{s\left( {s - b} \right)}}} = \frac{1}{3} \cr & \Rightarrow \,\,\frac{{s - c}}{s} = \frac{1}{3} \cr & \Rightarrow \,\,\frac{{a + b - c}}{{a + b + c}} = \frac{1}{3} \cr & \Rightarrow \,\,3a + 3b - 3c = a + b + c \cr & \Rightarrow \,\,a + b = 2c \cr & \Rightarrow \,\,a,b,c\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr} $$

We know by Sine rule
$$\eqalign{ & \frac{c}{{\sin C}} = 2R \cr & \Rightarrow \,\,C = 2R\sin C \cr & \Rightarrow \,\,C = 2R\,\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr & \Rightarrow \,\,{\text{Also }}\tan \frac{C}{2} = \frac{r}{{s - c}} \cr & \Rightarrow \,\,r = s - c\,\,\,\,\,\left( {\because \,\,\angle C = {{90}^ \circ }} \right) \cr & \Rightarrow \,\,a + b - c = 2r \cr & {\text{or }}2r + c = a + b \cr & {\text{or 2}}r + 2R = a + b\,\,\left( {{\text{Using }}C = 2R} \right) \cr} $$

Using $$\left( {m + n} \right)\cot \theta = n\cot \beta - m\cot \alpha ,$$      we get,
$$\eqalign{ & \left( {3 + 2} \right)\cot \angle CDA = 2\cot {30^ \circ } - 3\cot {60^ \circ } \cr & \Rightarrow \,\,\cot \angle CDA = \frac{{\sqrt 3 }}{5}. \cr} $$
Now use sine rule in $$\vartriangle CDA.$$

$$\eqalign{ & \tan \left( {\frac{\pi }{4}} \right) = \frac{a}{{2r}};\sin \left( {\frac{\pi }{n}} \right) = \frac{a}{{2R}} \cr & r + R = \frac{a}{2}\left[ {\cot \frac{\pi }{n} + {\text{cosec}}\frac{\pi }{n}} \right] \cr} $$

$$\eqalign{ & = \frac{a}{2}\left[ {\frac{{\cos \frac{\pi }{n} + 1}}{{\sin \frac{\pi }{n}}}} \right] \cr & = \frac{a}{2}\left[ {\frac{{2{{\cos }^2}\frac{\pi }{{2n}}}}{{2\sin \frac{\pi }{{2n}}\cos \frac{\pi }{{2n}}}}} \right] \cr & = \frac{a}{2}\cot \frac{\pi }{{2\pi }} \cr} $$

$$\eqalign{ & \frac{a}{{b + c}} = \frac{{\sin A}}{{\sin B + \sin C}} \cr & \frac{a}{{b + c}} = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\sin \frac{{B + C}}{2} \cdot \cos \frac{{B - C}}{2}}} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{{B - C}}{2}}} \cr & \therefore \,\,\sin \frac{A}{2} = \frac{a}{{b + c}}\cos \frac{{B - C}}{2} \leqslant \frac{a}{{b + c}}. \cr & \frac{{2a}}{{a + b + c}} = \frac{{2\sin A}}{{\sin A + \sin B + \sin C}} = \frac{{2 \cdot 2\sin \frac{A}{2}\cos \frac{A}{2}}}{{4\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}}} \cr & \frac{{2a}}{{a + b + c}} = \frac{{\sin \frac{A}{2}}}{{\cos \frac{B}{2}\cos \frac{C}{2}}} \geqslant \sin \frac{A}{2}. \cr} $$

Let $$a = \sin \alpha ,b = \cos \alpha \,\,{\text{and }}c = \sqrt {1 + \sin \alpha \cos \alpha } $$
Clearly $$a$$ and $$b < 1$$  but $$c > 1$$  as $$\sin \alpha > 0\,\,{\text{and}}\,\,\cos \alpha > 0$$
∴ $$c$$ is the greatest side and greatest angle is $$C$$
$$\eqalign{ & \therefore \,\,\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr & = \frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha }}{{2\sin \alpha \cos \alpha }} = - \frac{1}{2} \cr & \therefore \,\,C = {120^ \circ } \cr} $$

Let $$a = 3x + 4y, b = 4x + 3y$$      and $$c = 5x + 5y$$
as $$x, y > 0, c = 5x + 5y$$     is the largest side
∴ $$C$$ is the largest angle . Now
$$\eqalign{ & \cos C = \frac{{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}}}{{2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} \cr & = \frac{{ - 2xy}}{{2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0 \cr} $$
∴ $$C$$ is obtuse angle
⇒ $$\Delta ABC$$  is obtuse angled

$${\text{as }}\angle CPA = {45^ \circ }$$

$$\eqalign{ & {\text{so, }}AC = AP = x + 3 \cr & \tan {30^ \circ } = \frac{{AB}}{{AP}} = \frac{x}{{x + 3}} \cr & \frac{1}{{\sqrt 3 }} = \frac{x}{{x + 3}} \cr & x + 3 = \sqrt 3 \,x \cr & x = \frac{3}{{\sqrt 3 - 1}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & x = \frac{{3 \times 2.73}}{2} = \frac{{8.19}}{2} = 4.095\,m \approx 4.1\,m \cr} $$

$$\eqalign{ & \cos C = \frac{{{8^2} + {{10}^2} - {{12}^2}}}{{2 \cdot 8 \cdot 10}} = \frac{1}{8} \cr & \cos A = \frac{{{{10}^2} + {{12}^2} - {8^2}}}{{2 \cdot 10 \cdot 12}} = \frac{3}{4}. \cr & \cos 2A = 2{\cos ^2}A - 1 = 2 \times \frac{9}{{16}} - 1 = \frac{1}{8}. \cr & {\text{So, }}\cos C = \cos 2A \cr & \Rightarrow \,\,C = 2A. \cr} $$

Here, $$a + b = 2\sqrt 3 ,ab = 2\,{\text{and }}C = \frac{\pi }{3}.$$
$$\eqalign{ & \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr & \Rightarrow \,\,\frac{1}{2} = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr & \Rightarrow \,\,{a^2} + {b^2} - {c^2} = ab\,\,{\text{or, }}{\left( {a + b} \right)^2} - 2ab - {c^2} = ab \cr & {\text{or, }}12 - 4 - {c^2} = 2\,\,\,{\text{or, }}c = \sqrt 6 . \cr} $$
∴ the perimeter $$ = a + b + c = 2\sqrt 3 + \sqrt 6 .$$