$$\eqalign{ & \cos \frac{\pi }{4} = \frac{{{{\left( {2\sqrt 3 } \right)}^2} + {x^2} - {{\left( {2\sqrt 2 } \right)}^2}}}{{2 \cdot 2\sqrt 3 \cdot x}}\,\,{\text{or, }}\frac{{4\sqrt 3 }}{{\sqrt 2 }}x = {x^2} + 4\,\,{\text{or, }}{x^2} - \frac{{4\sqrt 3 }}{{\sqrt 2 }}x + 4 = 0. \cr & \therefore \,\,x = \sqrt 6 \pm \sqrt 2 . \cr} $$

$$\eqalign{ & {\text{From, }}\Delta \,ADB,AD = BD = x \cr & {\text{In }}\Delta \,ADC, \cr & \tan {30^ \circ } = \frac{x}{{\sqrt 3 + 1 - x}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{x}{{\sqrt 3 + 1 - x}} \cr & \Rightarrow \sqrt 3 \,x = \sqrt 3 + 1 - x \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)x = \sqrt 3 + 1 \cr & x = \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} \cr & {\text{Area of }}\Delta \,ABC = \frac{1}{2} \times \left( {\sqrt 3 + 1} \right) \times 1 = \frac{{\sqrt 3 + 1}}{2} \cr} $$

$$\eqalign{ & \tan {60^ \circ } = \frac{{150}}{x} \cr & \Rightarrow x = \frac{{150}}{{\sqrt 3 }} \cr & {\text{Also, }}\tan {45^ \circ } = \frac{{150}}{{x + y}} \cr & \Rightarrow x + y = 150 \cr & \Rightarrow y = 150 - x = 150 - \frac{{150}}{{\sqrt 3 }} \cr & \Rightarrow y = 150\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) = {\text{distance travelled}} \cr & {\text{Speed in}}\left( {m/hr} \right) = \frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{60}}{2} \cr & = 4500\frac{{\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & {R_1} = \frac{{BC}}{{2\sin \left( {\angle BOC} \right)}} = \frac{a}{{2\sin 2A}} \cr & \therefore \frac{a}{{{R_1}}} = 2\sin 2A \cr & {\text{Similarly}},\,\,\frac{b}{{{R_2}}} = 2\sin 2B \cr & {\text{and }}\frac{c}{{{R_3}}} = 2\sin 2C \cr & {\text{So}},\,\,\frac{a}{{{R_1}}} + \frac{b}{{{R_2}}} + \frac{c}{{{R_3}}} \cr & = 2\left( {\sin 2A + \sin 2B + \sin 2C} \right) \cr & = 2 \cdot 4\sin A\sin B\sin C\,\,\,\left[ {\because A + B + C = \pi } \right] \cr & = \left( {2\sin A} \right)\left( {2\sin B} \right)\left( {2\sin C} \right) = \left( {\frac{a}{R}} \right)\left( {\frac{b}{R}} \right)\left( {\frac{c}{R}} \right) \cr} $$

$$\eqalign{ & \frac{{2R\sin A}}{{\cos A}} = \frac{{2R\sin B}}{{\cos B}}\,\,{\text{or, }}\sin A\cos B = \cos A\sin B \cr & {\text{or, }}2\sin A\cos B = \cos A\sin B + \sin A\cos B = \sin \left( {A + B} \right) = \sin C. \cr} $$

Let $$OP =$$  Pole,
$$\eqalign{ & \angle PAO = \angle PBO = \angle PCQ = \alpha \cr & \frac{{OP}}{{OB}} = \tan \alpha \cr & \Rightarrow \,\,OB = OP\cot \alpha \,\,\,\,\,\,.....\left( 1 \right) \cr} $$
$$\eqalign{ & {\text{Similarly}}\,\,OA = OP\cot \alpha \,\,\,\,\,.....\left( 2 \right) \cr & {\text{Similarly}}\,\,OC = OP\cot \alpha \,\,\,\,\,.....\left( 3 \right) \cr} $$
From (1), (2) and (3), $$OA = OB = OC$$
⇒ $$O$$ is the point of circumcentre of the triangle $$ABC.$$

$$\eqalign{ & {\text{We have, }}\,\tan A = \frac{a}{{2{d_1}}}; \cr & {d_1} = R\cos A\,{\text{ etc}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Similarly, }}\,\tan B = \frac{b}{{2{d_2}}} \cr & {\text{and }}\,\tan C = \frac{C}{{2{d_3}}} \cr & {\text{In }}\,\Delta \,ABC,\tan A + \tan B + \tan C \cr & = \tan A \cdot \tan B \cdot \tan C \cr & \Rightarrow \frac{a}{{2{d_1}}} + \frac{b}{{2{d_2}}} + \frac{c}{{2{d_3}}} = \frac{{abc}}{{8{d_1}{d_2}{d_3}}} \cr & \therefore 4\left( {\frac{a}{{{d_1}}} + \frac{b}{{{d_2}}} + \frac{c}{{{d_3}}}} \right) = \frac{{abc}}{{{d_1}{d_2}{d_3}}} \cr & \Rightarrow \lambda = 4 \cr} $$

$$\eqalign{ & \sqrt 3 \cos A + \sin A = \sqrt 3 \cos B + \sin B \cr & {\text{or, }}\frac{{\sin A - \sin B}}{{\cos B - \cos A}} = \sqrt 3 \cr & \therefore \,\,\sqrt 3 = \frac{{2\cos \frac{{A + B}}{2} \cdot \sin \frac{{A - B}}{2}}}{{2\sin \frac{{A + B}}{2} \cdot \sin \frac{{A - B}}{2}}} = \cot \frac{{A + B}}{2} \cr & {\text{or, }}\frac{{A + B}}{2} = \frac{\pi }{6}\,\,\,{\text{or, }}A + B = \frac{\pi }{3} \cr & \therefore \,\,C = \pi - \frac{\pi }{3}. \cr} $$

$$\eqalign{ & 2\Delta = \left( {a - b + c} \right)\left( {a + b - c} \right) \cr & \Rightarrow 2\Delta = 4\left( {s - b} \right)\left( {s - c} \right) \cr & \Rightarrow 2 = \sqrt {\frac{{s\left( {s - a} \right)}}{{\left( {s - b} \right)\left( {s - c} \right)}}} \cr & \Rightarrow \tan \frac{A}{2} = \frac{1}{2} \cr & \therefore \tan A = \frac{{2 \times \frac{1}{2}}}{{1 - \frac{1}{4}}} = \frac{4}{3} \cr} $$

Sides are in the Ration $$1:\sqrt 3 :2$$
$$\eqalign{ & {\text{Let }}a = k,b = \sqrt 3 k\,\,{\text{and }}c = 2k \cr & \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{\sqrt 3 }}{2} \cr & \Rightarrow \,\,A = \frac{\pi }{6} \cr & \cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \frac{1}{2} \cr & \Rightarrow \,\,B = \frac{\pi }{3} \cr & \Rightarrow \,\,C = \pi - \left( {A + B} \right) = \frac{\pi }{2} \cr & \Rightarrow \,\,A:B:C = 1:2:3 \cr} $$