$$\eqalign{ & {r_1} > {r_2} > {r_3} \cr & \Rightarrow \,\,\frac{\Delta }{{s - a}} > \frac{\Delta }{{s - b}} > \frac{\Delta }{{s - c}}; \cr & \Rightarrow \,\,s - a < s - b < s - c \cr & \Rightarrow \,\, - a < - b < - c \cr & \Rightarrow \,\,a > b > c \cr} $$

$$\eqalign{ & \frac{{\sin {{38}^ \circ }}}{l} = \frac{{\sin \left( {SPO} \right)}}{{2.05}} \cr & = \frac{{\sin \left( {{{180}^ \circ } - {{38}^ \circ } - {{90}^ \circ } - {{10}^ \circ }} \right)}}{{2.05}} \cr & \Rightarrow l = \frac{{2.05\sin {{38}^ \circ }}}{{\sin {{42}^ \circ }}} \cr} $$

$$b = ar,c = a{r^2},$$    where $$r > 1.$$
From the question, $$C = 2A.\,\,{\text{So}},B = \pi - A - C = \pi - 3A.$$
$$\eqalign{ & \therefore \,\,\frac{a}{{\sin \,A}} = \frac{b}{{\sin \,B}} = \frac{c}{{\sin \,C}} \cr & \Rightarrow \,\,\frac{1}{{\sin \,A}} = \frac{r}{{\sin \,3A}} = \frac{{{r^2}}}{{\sin \,2A}} \cr & \therefore \,\,{r^2} = 2\cos \,A\,\,{\text{and}}\,\,r = \frac{{\sin \,3A}}{{\sin \,A}} = 3 - 4{\sin ^2}A = 4{\cos ^2}A - 1 \cr & \therefore \,\,r = {r^4} - 1\,\,\,{\text{or,}}\,\,{r^4} = 1 + r. \cr} $$
Among $$\sqrt 3 ,\root 4 \of 3 ,\frac{{\sqrt 5 + 1}}{2}$$    we find $$\root 4 \of 3 $$ is the smallest because $${\left( {\sqrt 3 } \right)^4} = 9,{\left( {\root 4 \of 3 } \right)^4} = 3,{\left( {\frac{{\sqrt 5 + 1}}{2}} \right)^4} = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} = \frac{{14 + 6\sqrt 5 }}{4} > 3.$$
Now, putting $$r = \root 4 \of 3 ,$$  we have $$3 > 1 + \root 4 \of 3 ,$$   implying $${\root 4 \of 3 }$$ does not satisfy $${r^4} = 1 + r.\,\,{\text{So, }}r < \root 4 \of 3 .$$

$${\text{Given }}AB\parallel CD,$$   $$CD = 2AB$$
Let $$AB = a$$   then $$CD = 2a$$   Let radius of circle be $$r.$$ Let circle touches $$AB$$  at $$P, BC$$  at $$Q, AD$$  at $$R$$ and $$CD$$  at $$S.$$
Then $$AR = AP = r, BP = BQ = a - r$$
$$DR = DS = r$$   and $$CQ = CS = 2a - r$$     In $$\Delta BEC$$
$$\eqalign{ & B{C^2} = B{E^2} + E{C^2} \cr & \Rightarrow \,\,{\left( {a - r + 2a - r} \right)^2} = {\left( {2r} \right)^2} + {\left( a \right)^2} \cr & \Rightarrow \,\,9{a^2} + 4{r^2} - 12ar = 4{r^2} + {a^2} \cr & \Rightarrow \,\,a = \frac{3}{2}r\,\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
Also $$Ar$$ (quad. $$ABCD$$   ) $$=$$ 18

$$\eqalign{ & \Rightarrow \,\,a \times 2r + \frac{1}{2} \times a \times 2r = 18 \cr & \Rightarrow \,\,ar = 6 \cr & \Rightarrow \,\,\frac{{3{r^2}}}{2} = 6\,\left( {{\text{using equation }}\left( 1 \right)} \right) \cr & \Rightarrow \,\,{r^2} = 4 \cr & \Rightarrow \,\,r = 2 \cr} $$

$$\frac{1}{2} \cdot p \cdot a = {\text{area}} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} .$$

$$\eqalign{ & {d_2} = h\cot {30^ \circ } = 500\sqrt 3 , \cr & {\text{and }}{d_1} = \frac{{500}}{{\sqrt 3 }} \cr} $$

$${\text{Diameter }}d = 500\sqrt 3 + \frac{{500}}{3}\sqrt 3 = \frac{{2000}}{{\sqrt 3 }}m$$

$$\eqalign{ & \tan\frac{{A - B}}{2} = \frac{1}{3}\tan \frac{{A + B}}{2} = \frac{1}{3}\cot \frac{C}{2} \cr & \Rightarrow \,\,\frac{{a - b}}{{a + b}} = \frac{1}{3}. \cr} $$

$$\eqalign{ & \tan \frac{A}{2} + \tan \frac{B}{2} = \frac{5}{6},\tan \frac{A}{2} \cdot \tan \frac{B}{2} = \frac{1}{6} \cr & \Rightarrow \,\,\tan \left( {\frac{{A + B}}{2}} \right) = 1 \cr & \Rightarrow \,\,A + B = 2 \times \frac{\pi }{4}. \cr} $$

Height of tree is
$$AB + AC = \frac{{20}}{{\sqrt 3 }} + \frac{{10}}{{\sqrt 3 }} = \frac{{30}}{{\sqrt 3 }} = 10\sqrt 3 = 17.32\,m$$

$$\left( {\alpha - \beta } \right) + \alpha + \left( {\alpha + \beta } \right) = {180^ \circ }\,\,{\text{and }}\alpha + \beta = {90^ \circ }.$$
∴ the angles are $${30^ \circ },{60^ \circ },{90^ \circ }.$$
$$\eqalign{ & \therefore \,\,\frac{a}{{\sin {{30}^ \circ }}} = \frac{b}{{\sin {{60}^ \circ }}} = \frac{c}{{\sin {{90}^ \circ }}} = 2R\,\,\,{\text{or, }}a = R,b = \sqrt 3 R,c = 2R. \cr & r = \frac{\vartriangle }{s} = \frac{{\frac{1}{2}R \cdot \sqrt 3 R}}{{\frac{1}{2}\left( {R + \sqrt 3 R + 2R} \right)}} = \frac{{\sqrt 3 R}}{{3 + \sqrt 3 }} \cr & {\text{and, }}2s = R + \sqrt 3 R + 2R = \left( {3 + \sqrt 3 } \right)R. \cr & \therefore \,\,\frac{r}{{2s}} = \frac{{\sqrt 3 R}}{{3 + \sqrt 3 }} \times \frac{1}{{\left( {3 + \sqrt 3 } \right)R}} = \frac{1}{{\left( {3 + \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)}} \cr & \frac{r}{{2s}} = \frac{1}{{6 + 4\sqrt 3 }} = \frac{1}{{2\sqrt 3 \left( {2 + \sqrt 3 } \right)}} = \frac{{2 - \sqrt 3 }}{{2\sqrt 3 }}. \cr} $$