$$\eqalign{ & \cot A \cdot \cot B \cdot \cot C > 0 \cr & \Rightarrow \,\,\cot A > 0,\cot B > 0,\cot C > 0 \cr} $$
because two or more of $$\cot A, \cot B, \cot C$$    cannot be negative at the same time in a triangle

If $$O$$ is centre of polygon and $$AB$$  is one of the side, then by figure $$\cos \frac{\pi }{n} = \frac{r}{R}$$

$$ \Rightarrow \,\,\frac{r}{R} = \frac{1}{2},\frac{1}{{\sqrt 2 }},\frac{{\sqrt 3 }}{2}\,\,{\text{for}}$$
$$n = 3, 4, 6$$   respectively.

$$OA = OB = h\cot \theta $$

Now,
$$\eqalign{ & AB = 2AN = 2AO\sin \frac{\theta }{2} \cr & = 2h\frac{{\cos \theta }}{{\sin \theta }} \cdot \sin \frac{\theta }{2} = h\cos \theta \sec \frac{\theta }{2} \cr} $$

$$\eqalign{ & {\text{In }}\vartriangle ABC,\cos 2A + \cos 2B - \cos 2C = 1 - 4\sin A\sin B\cos C < 1\,\,{\text{for all }}A,B. \cr & \cos 2A + \cos 2B + \cos 2C = - 1 - 4\cos A\cos B\cos C < - 1\,\,{\text{for acute }}A,B. \cr & {\cos ^2}A + {\cos ^2}B + {\cos ^2}C = 1 - 2\cos A\cos B\cos C < 1\,\,{\text{for acute }}A,B. \cr} $$

Clearly, $$\angle BOD = 2C$$
$$\eqalign{ & \therefore \,\,\cos 2C = \frac{{{1^2} + {1^2} - {{\left( {\sqrt 3 } \right)}^2}}}{{2 \cdot 1 \cdot 1}}\,\,{\text{or, }}\cos 2C = - \frac{1}{2} \cr & \therefore \,\,C = {60^ \circ } \cr & \therefore \,\,A = {180^ \circ } - C = {120^ \circ }. \cr} $$

So, $$\cos{120^ \circ } = \frac{{A{D^2} + {1^2} - {{\left( {\sqrt 3 } \right)}^2}}}{{2 \cdot AD \cdot 1}}$$
$$\eqalign{ & \therefore \,\, - \frac{1}{2} = \frac{{A{D^2} - 2}}{{2AD}}\,\,{\text{or, }}A{D^2} + AD - 2 = 0 \cr & {\text{or, }}AD = 1. \cr} $$
Now, $${\text{ar}}\left( {ABCD} \right) = {\text{ar}}\left( {\vartriangle ADB} \right) + {\text{ar}}\left( {\vartriangle BCD} \right)$$
or, $$\frac{{3\sqrt 3 }}{4} = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin {120^ \circ } + \frac{1}{2} \cdot BC \cdot CD \cdot \sin {60^ \circ }\,\,{\text{or, }}BC \cdot CD = 2.$$

$$\eqalign{ & \cos {120^ \circ } = \frac{{{x^2} + {x^2} - A{B^2}}}{{2{x^2}}} \cr & \Rightarrow \frac{{2{x^2} - A{B^2}}}{{2{x^2}}} = \frac{{ - 1}}{2} \cr & \Rightarrow 4{x^2} - 2A{B^2} = - 2{x^2} \cr} $$

$$\eqalign{ & \Rightarrow 3{x^2} = A{B^2} \cr & \Rightarrow AB = x\sqrt 3 \cr & \Rightarrow {a^2}:{b^2}:{c^2} = {\left( {2x} \right)^2}:{x^2}:{\left( {x\sqrt 3 } \right)^2} \cr & = 4{x^2}:{x^2}:3{x^2} = 4:1:3 \cr} $$

$$\eqalign{ & 3\sin \,A = \sin \,B + \sin \,C \cr & \Rightarrow \,\,6\sin \frac{A}{2} \cdot \cos \frac{A}{2} \cdot = 2\sin \frac{{B + C}}{2} \cdot \cos \frac{{B - C}}{2} \cr & \Rightarrow \,\,3\cos \frac{{B + C}}{2} = \cos \frac{{B - C}}{2} \cr & \Rightarrow \,\,3\left( {\cos \frac{B}{2} \cdot \cos \frac{C}{2} - \sin \frac{B}{2} \cdot \sin \frac{C}{2}} \right) = \cos \frac{B}{2} \cdot \cos \frac{C}{2} + \sin \frac{B}{2}\sin \frac{C}{2} \cr & \Rightarrow \,\,2\cos \frac{B}{2} \cdot \cos \frac{C}{2} = 4\sin \frac{B}{2} \cdot \sin \frac{C}{2} \cr & \Rightarrow \,\,\tan \frac{B}{2} \cdot \tan \frac{C}{2} = \frac{1}{2}. \cr} $$

Let height of hill $$= H$$
& horizontal distance between building & hill $$= d$$
$$\eqalign{ & \tan q = \frac{H}{d} \cr & \Rightarrow d = \frac{H}{{\tan q}} = H\cot q \cr & \tan p = \frac{{\left( {H - h} \right)}}{d} \cr & \Rightarrow d = \left( {H - h} \right)\cot p \cr & \Rightarrow H\cot q = \left( {H - h} \right)\cot p \cr & H = \frac{{h\cot p}}{{\cot p - \cot q}} \cr} $$

Here, $$2b = a + c$$   and $$\vartriangle = \frac{3}{5} \times \frac{{\sqrt 3 }}{4} \cdot {\left( {\frac{{a + b + c}}{3}} \right)^2}\,\,\,\therefore \,\,\vartriangle = \frac{{3\sqrt 3 }}{{20}} \cdot {b^2}.$$
$$\eqalign{ & \therefore \,\,\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} = \frac{{3\sqrt 3 }}{{20}}{b^2} \cr & {\text{or, }}\sqrt {\frac{1}{{16}}\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)} = \frac{{3\sqrt 3 }}{{20}}{b^2} \cr & {\text{or, }}\sqrt {3b \cdot \left( {b + c + c - 2b} \right)b\left( {2b - c + b - c} \right)} = \frac{{3\sqrt 3 }}{5}{b^2} \cr & {\text{or, }}\sqrt {\left( {2c - b} \right)\left( {3b - 2c} \right)} = \frac{3}{5}b\,\,\,{\text{or, }}8bc - 4{c^2} - 3{b^2} = \frac{9}{{25}}{b^2} \cr & {\text{or, }}\frac{{84}}{{25}}{b^2} - 8bc + 4{c^2} = 0\,\,\,\,\,{\text{or, }}\frac{b}{c} = \frac{{8 \pm \sqrt {64 - 16 \cdot \frac{{84}}{{25}}} }}{{2 \cdot \frac{{84}}{{25}}}} = \frac{5}{7},\frac{5}{3}. \cr & {\text{Also, }}2b = a + c \cr & \Rightarrow \,\,2\frac{b}{c} = \frac{a}{c} + 1 \cr & \Rightarrow \,\,\frac{a}{c} = \frac{{2b}}{c} - 1 = \frac{3}{7},\frac{7}{3}. \cr} $$

$$\eqalign{ & {a^2} + {a^2} + 4{b^2} - 4ab = 2ac - {c^2} \cr & \Rightarrow {\left( {a - 2b} \right)^2} + {\left( {a - c} \right)^2} = 0 \cr} $$
which is possible only when : $$a - 2b =0$$   and $$a - c = 0$$
$$\eqalign{ & {{\text{or }}} \frac{a}{1} = \frac{b}{{\frac{1}{2}}} = \frac{c}{1} = \lambda \left( {{\text{say}}} \right) \cr & \therefore a = \lambda ,b = \frac{\lambda }{2},c = \lambda \cr & \therefore \cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}} \cr & = \frac{{{\lambda ^2} + {\lambda ^2} - \frac{{{\lambda ^2}}}{4}}}{{2{\lambda ^2}}} = 1 - \frac{1}{8} = \frac{7}{8} \cr} $$