$$\eqalign{ & \frac{1}{{{r_1}}},\frac{1}{{{r_2}}},\frac{1}{{{r_3}}}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,\frac{{s - a}}{\vartriangle },\frac{{s - b}}{\vartriangle },\frac{{s - c}}{\vartriangle }\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\,s - a,s - b,s - c\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow \,\, - a, - b, - c\,{\text{are in A}}{\text{.P}}{\text{.}} \cr} $$

$$\frac{1}{2}\alpha \cdot a = \vartriangle ;$$
$$\therefore \,\,\frac{1}{\alpha } = \frac{a}{{2\vartriangle }}.$$   Similarly for others.
$$\therefore \,\,{\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}} = \frac{1}{{2\vartriangle }}\left( {a + b + c} \right) = \frac{s}{\vartriangle }.$$

In the $$\vartriangle OAB,OA = OB = r\,\,{\text{and }}\angle AOB = \frac{{{{360}^ \circ }}}{5} = {72^ \circ }.$$

$$\eqalign{ & \therefore \,\,{\text{ar}}\left( {\vartriangle AOB} \right) = \frac{1}{2} \cdot r \cdot r \cdot \sin {72^ \circ } = \frac{1}{2}{r^2}\cos {18^ \circ } \cr & \therefore \,\,{\text{ar}}\left( {{\text{pentagon}}} \right) = \frac{5}{2}{r^2}\cos {18^ \circ } \cr & \therefore \,\,{A_1}:{A_2} = \frac{{2\pi {r^2}}}{{5{r^2}\cos {{18}^ \circ }}} = \frac{{2\pi }}{5}\sec \frac{\pi }{{10}}. \cr} $$

Let $$h$$ be the height of tree $$PQ$$  and breadth of river $$PS$$  be $$x \,ft.$$
Angle of elevation subtended by a tree is $${60^ \circ }.$$
Also, when he retreats 20 feet, the angle becomes $${30^ \circ }.$$

$$\eqalign{ & {\text{Also, in }}\,\Delta \,PQS,\tan {60^ \circ } = \frac{h}{x} \cr & \Rightarrow h = \sqrt 3 x \cr & {\text{and in }}\,\Delta \,PQR,\tan {30^ \circ } = \frac{h}{{x + 20}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + 20}} \cr & \Rightarrow x + 20 = \sqrt 3 h \cr & \Rightarrow x + 20 = 3x\,\left( {{\text{By putting value of }}h} \right) \cr & \Rightarrow 2x = 20 \cr & \Rightarrow x = 10 \cr} $$
Hence breadth of river is $$10\,ft.$$

$$\because \,\,PB\,{\text{bisects }}\angle APC,$$     therefore $$AB : BC = PA : PC$$
Also in $$\Delta \,APQ,\sin{30^ \circ } = \frac{h}{{PA}}$$
$$ \Rightarrow \,\,PA = 2h$$
and in $$\Delta \,CPQ,\sin {60^ \circ } = \frac{h}{{PC}}$$
$$ \Rightarrow \,\,PC = \frac{{2h}}{{\sqrt 3 }}$$
$$\therefore \,\,AB:BC = 2h:\frac{{2h}}{{\sqrt 3 }} = \sqrt 3 :1$$

$$\eqalign{ & \because \,\,A,B,C\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,B = {60^ \circ }\,\,{\text{then}}\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A \cr & = \frac{{K\sin A}}{{K\sin C}}.2\sin C\cos C + \frac{{K\sin C}}{{K\sin A}},\sin A\cos A \cr & = 2\left( {\sin A\cos C + \cos A\sin C} \right) \cr & = 2\sin \left( {A + C} \right) \cr & = 2\sin B \cr & = 2 \times \frac{{\sqrt 3 }}{2} \cr & = \sqrt 3 \cr} $$

$$\eqalign{ & \frac{{\cot A + \cot B}}{{\cot C}} = \frac{{\sin \left( {A + B} \right)}}{{\sin A\sin B}} \cdot \frac{{\sin C}}{{\sin C}} \cr & = \frac{{{{\sin }^2}C}}{{\sin A\sin B\cos C}} = \frac{{{c^2}}}{{ab}} \cdot \frac{{2ab}}{{{a^2} + {b^2} - {c^2}}} \cr & = \frac{{2{c^2}}}{{{a^2} + {b^2} - {c^2}}} = \frac{{2{c^2}}}{{\frac{{17{c^2}}}{9} - {c^2}}} = \frac{9}{4} = \frac{m}{n} \cr & \Rightarrow \left( {m + n} \right) = 9 + 4 = 13 \cr} $$

$$\eqalign{ & {\text{In }}\,\Delta \,ABC,\tan {45^ \circ } = \frac{{AB}}{{AC}} = \frac{h}{x};1 = \frac{h}{x} \cr & h = x\,\,\,.....\left( {\text{i}} \right) \cr & {\text{In }}\Delta \,ABD, \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}};\,\,\,\frac{1}{{\sqrt 3 }} = \frac{h}{{x + 20}} \cr & x + 20 = \sqrt 3 h;\,\,\,h + 20 = \sqrt 3 h \cr & 20 = \left( {\sqrt 3 - 1} \right)h;\,\,\,h = \frac{{20}}{{\sqrt 3 - 1}} \cr & = \frac{{20}}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} \cr & = \frac{{20\left( {\sqrt 3 + 1} \right)}}{2} = 10\left( {\sqrt 3 + 1} \right)m \cr} $$
Hence, the height is $$10\left( {\sqrt 3 + 1} \right)m$$

$$\eqalign{ & \cos A\sin B = \sin C \cr & \Rightarrow \sin \left( {A + B} \right) - \sin \left( {A - B} \right) = 2\sin C \cr & \Rightarrow \sin C = \sin \left( {B - A} \right) \cr & \Rightarrow A + C = B\,\left( {\because A + B = \pi - C} \right) \cr & \therefore B = \frac{\pi }{2} \cr & {\text{Now, }}3b - 5c = 0 \cr & \Rightarrow 3 - 5\sin C = 0 \cr & \therefore \sin C = \frac{3}{5}\,{\text{and}}\,A = \frac{\pi }{2} - C \cr & \Rightarrow \cos A = \sin C \cr & \Rightarrow \frac{{1 - {{\tan }^2}\frac{A}{2}}}{{1 + {{\tan }^2}\frac{A}{2}}} = \frac{3}{5} \cr & \therefore {\tan ^2}\frac{A}{2} = \frac{1}{4} \cr & \Rightarrow \tan \frac{A}{2} = 0.5 \cr} $$

We know that, $$2s = a + b + c$$
$$\eqalign{ & \therefore \frac{{\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)}}{{4{b^2}{c^2}}} \cr & = \frac{{2s\left( {2s - 2a} \right)\left( {2s - 2b} \right)\left( {2s - 2c} \right)}}{{4{b^2}{c^2}}} \cr & = 4\frac{{s\left( {s - a} \right)}}{{bc}} \times \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}} \cr & = 4{\cos ^2}\frac{A}{2} \times {\sin ^2}\frac{A}{2} \cr & = {\sin ^2}A \cr} $$