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Question-88683




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 2  D(2(√2),−(√2))    eqn. of AD:  (y/(x+3))=((−(√2))/(2(√2)+3))  y=−(3(√2)−4)(x+3)    eqn. of OC:  y=−x    y_B =−3(3(√2)−4)  a=(√(3^2 +3^2 (3(√2)−4)^2 ))=3(√(35−24(√2)))    y_C =−(3(√2)−4)(x_C +3)=−x_C   x_C =((3(3(√2)−4))/(5−3(√2)))=((3(3(√2)−2))/7)  y_C =−((3(3(√2)−2))/7)  b=(√((2(√2)−((3(3(√2)−2))/7))^2 +(−(√2)+((3(3(√2)−2))/7))^2 ))=((√(130+36(√2)))/7)    (a/b)=((21(√(35−24(√2))))/( (√(130+36(√2)))))=1.6066
$${WAY}\:\mathrm{2} \\ $$$${D}\left(\mathrm{2}\sqrt{\mathrm{2}},−\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{AD}: \\ $$$$\frac{{y}}{{x}+\mathrm{3}}=\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}} \\ $$$${y}=−\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{OC}: \\ $$$${y}=−{x} \\ $$$$ \\ $$$${y}_{{B}} =−\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right) \\ $$$${a}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{35}−\mathrm{24}\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$${y}_{{C}} =−\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)\left({x}_{{C}} +\mathrm{3}\right)=−{x}_{{C}} \\ $$$${x}_{{C}} =\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{5}−\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}} \\ $$$${y}_{{C}} =−\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}} \\ $$$${b}=\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}}\right)^{\mathrm{2}} +\left(−\sqrt{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}{\mathrm{7}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{130}+\mathrm{36}\sqrt{\mathrm{2}}}}{\mathrm{7}} \\ $$$$ \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{21}\sqrt{\mathrm{35}−\mathrm{24}\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{130}+\mathrm{36}\sqrt{\mathrm{2}}}}=\mathrm{1}.\mathrm{6066} \\ $$
Commented by jagoll last updated on 14/Apr/20
waw...not exact sir. thank you
$${waw}…{not}\:{exact}\:{sir}.\:{thank}\:{you} \\ $$
Answered by mr W last updated on 12/Apr/20
Commented by mr W last updated on 12/Apr/20
AD=1+3+3(√2)=4+3(√2)  GD=3(√2)+3  AG=(√((4+3(√2))^2 +(3+3(√2))^2 −2(4+3(√2))(3+3(√2))×((√2)/2)))=(√(19+12(√2)))  ((sin ∠DAG)/(3+3(√2)))=((sin ∠DGA)/(4+3(√2)))=((sin 45°)/( (√(19+12(√2)))))  ⇒sin ∠DAG=((3+3(√2))/( (√(2(19+12(√2))))))=((3(2+(√2)))/(2(√(19+12(√2)))))  ⇒sin ∠DGA=((4+3(√2))/( (√(2(19+12(√2))))))=((3+2(√2))/( (√(19+12(√2)))))  b=(1/(cos ∠DAG))=1.921  a=(3/(sin ∠GDA))=3.087  (a/b)=1.6066
$${AD}=\mathrm{1}+\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}=\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${GD}=\mathrm{3}\sqrt{\mathrm{2}}+\mathrm{3} \\ $$$${AG}=\sqrt{\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}\right)\left(\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}\right)×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{sin}\:\angle{DAG}}{\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{sin}\:\angle{DGA}}{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}=\frac{\mathrm{sin}\:\mathrm{45}°}{\:\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{DAG}=\frac{\mathrm{3}+\mathrm{3}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\left(\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}\right)}}=\frac{\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow\mathrm{sin}\:\angle{DGA}=\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\left(\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}\right)}}=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{19}+\mathrm{12}\sqrt{\mathrm{2}}}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{cos}\:\angle{DAG}}=\mathrm{1}.\mathrm{921} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{sin}\:\angle{GDA}}=\mathrm{3}.\mathrm{087} \\ $$$$\frac{{a}}{{b}}=\mathrm{1}.\mathrm{6066} \\ $$

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