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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

$$WAY2$$$$D(2\sqrt{2},-\sqrt{2})$$$$$$$$eqn.ofAD:$$$$\frac{y}{x+3}=\frac{-\sqrt{2}}{2\sqrt{2}+3}$$$$y=-(3\sqrt{2}-4)(x+3)$$$$$$$$eqn.ofOC:$$$$y=-x$$$$$$$$y_B=-3(3\sqrt{2}-4)$$$$a=\sqrt{3^2+3^2{(3\sqrt{2}-4)}^2}=3\sqrt{35-24\sqrt{2}}$$$$$$$$y_C=-(3\sqrt{2}-4)(x_C+3)=-x_C$$$$x_C=\frac{3(3\sqrt{2}-4)}{5-3\sqrt{2}}=\frac{3(3\sqrt{2}-2)}{7}$$$$y_C=-\frac{3(3\sqrt{2}-2)}{7}$$$$b=\sqrt{{(2\sqrt{2}-\frac{3(3\sqrt{2}-2)}{7})}^2+{(-\sqrt{2}+\frac{3(3\sqrt{2}-2)}{7})}^2}=\frac{\sqrt{130+36\sqrt{2}}}{7}$$$$$$$$\frac{a}{b}=\frac{21\sqrt{35-24\sqrt{2}}}{\sqrt{130+36\sqrt{2}}}=1.6066$$

Commented by jagoll last updated on 14/Apr/20

$$waw...notexactsir.thankyou$$

Answered by mr W last updated on 12/Apr/20

Commented by mr W last updated on 12/Apr/20

$$AD=1+3+3\sqrt{2}=4+3\sqrt{2}$$$$GD=3\sqrt{2}+3$$$$AG=\sqrt{{(4+3\sqrt{2})}^2+{(3+3\sqrt{2})}^2-2(4+3\sqrt{2})(3+3\sqrt{2})\times \frac{\sqrt{2}}{2}}=\sqrt{19+12\sqrt{2}}$$$$\frac{\sin \mathrm{\angle }DAG}{3+3\sqrt{2}}=\frac{\sin \mathrm{\angle }DGA}{4+3\sqrt{2}}=\frac{\sin 45°}{\sqrt{19+12\sqrt{2}}}$$$$\Rightarrow \sin \mathrm{\angle }DAG=\frac{3+3\sqrt{2}}{\sqrt{2(19+12\sqrt{2})}}=\frac{3(2+\sqrt{2})}{2\sqrt{19+12\sqrt{2}}}$$$$\Rightarrow \sin \mathrm{\angle }DGA=\frac{4+3\sqrt{2}}{\sqrt{2(19+12\sqrt{2})}}=\frac{3+2\sqrt{2}}{\sqrt{19+12\sqrt{2}}}$$$$b=\frac{1}{\cos \mathrm{\angle }DAG}=1.921$$$$a=\frac{3}{\sin \mathrm{\angle }GDA}=3.087$$$$\frac{a}{b}=1.6066$$

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