Question Number 88683 by jagoll last updated on 12/Apr/20
Answered by mr W last updated on 12/Apr/20
Commented by mr W last updated on 12/Apr/20
$${WAY}\:\mathrm{2} \\ $$$${D}\left(\mathrm{2}\sqrt{\mathrm{2}},−\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{AD}: \\ $$$$\frac{{y}}{{x}+\mathrm{3}}=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$${y}=−\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{OC}: \\ $$$${y}=−{x} \\ $$$$ \\ $$$${y}_{{B}} =−\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right) \\ $$$${a}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{35}−\mathrm{24}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$${y}_{{C}} =−\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\left({x}_{{C}} +\mathrm{3}\right)=−{x}_{{C}} \\ $$$${x}_{{C}} =\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{5}−\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}} \\ $$$${y}_{{C}} =−\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}} \\ $$$${b}=\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}}\right)^{\mathrm{2}} +\left(−\sqrt{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{130}+\mathrm{36}\sqrt{\mathrm{2}}}}{\mathrm{7}} \\ $$$$ \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{21}\sqrt{\mathrm{35}−\mathrm{24}\sqrt{\mathrm{2}}}}{\sqrt{\mathrm{130}+\mathrm{36}\sqrt{\mathrm{2}}}}=\mathrm{1}.\mathrm{6066} \\ $$
Commented by jagoll last updated on 14/Apr/20
$${waw}...{not}\:{exact}\:{sir}.\:{thank}\:{you} \\ $$
Answered by mr W last updated on 12/Apr/20
Commented by mr W last updated on 12/Apr/20
$${AD}=\mathrm{1}+\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}=\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${GD}=\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{3} \\ $$$${AG}=\sqrt{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\left(\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}\right)×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{sin}\:\angle{DAG}}{\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{sin}\:\angle{DGA}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{sin}\:\mathrm{45}°}{\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{DAG}=\frac{\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}\left(\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}\right)}}=\frac{\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{DGA}=\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}\left(\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}\right)}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{cos}\:\angle{DAG}}=\mathrm{1}.\mathrm{921} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{sin}\:\angle{GDA}}=\mathrm{3}.\mathrm{087} \\ $$$$\frac{{a}}{{b}}=\mathrm{1}.\mathrm{6066} \\ $$