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Question Number 176816 by Ar Brandon last updated on 27/Sep/22

Commented by MJS_new last updated on 27/Sep/22

$$y=x\wedge 5x+12y=60\Rightarrow x=y=\frac{60}{17}$$

Commented by Ar Brandon last updated on 27/Sep/22

Oh! Thank you Sir MJS

Commented by cortano1 last updated on 27/Sep/22

$$\mathrm{D}(0,5),\mathrm{C}(5,5),\mathrm{B}(12,0),\mathrm{A}(0,0)$$$$\mathrm{AC}:\mathrm{y}=\mathrm{x};\mathrm{BD}:5\mathrm{x}+12\mathrm{y}=60$$$$\mathrm{Let}\mathrm{F}(\frac{60}{17},\frac{60}{17})$$$$\alpha ={\alpha }_1+{\alpha }_2\Rightarrow \sin \alpha =\sin ({\alpha }_1+{\alpha }_2)$$$$=\sin {\alpha }_1\cos {\alpha }_2+\cos {\alpha }_1\sin {\alpha }_2$$$$=\frac{1}{\sqrt{2}}.\frac{60}{156}+\frac{1}{\sqrt{2}}.\frac{144}{156}$$$$=\frac{1}{\sqrt{2}}\frac{204}{156}=\frac{17}{13\sqrt{2}}=\frac{17\sqrt{2}}{26}$$

Commented by Ar Brandon last updated on 27/Sep/22

$$\mathrm{Thanks}!$$$$\mathrm{But}\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{F}(\frac{60}{17},\frac{60}{17})?$$

Answered by mr W last updated on 27/Sep/22

$$AC=5\sqrt{2}$$$$DB=\sqrt{5^2+{12}^2}=13$$$$area=\frac{5\times (5+12)}{2}=\frac{5\sqrt{2}\times 13\times \sin \alpha }{2}$$$$\Rightarrow \sin \alpha =\frac{17}{13\sqrt{2}}=\frac{17\sqrt{2}}{26}$$

Answered by HeferH last updated on 27/Sep/22

$$$$$$\mathrm{\angle }DCA=45°$$$$\tan \left(\mathrm{\angle }DBA\right)=\frac{5}{12}$$$$\mathrm{\angle }DBA={\tan }^{-1}\left(\frac{5}{12}\right)\approx 22.7°$$$$a+45°+22.7°=180°$$$$a=112.3°$$$$\sin (112.3°)\approx 0.9$$$$$$

Commented by Ar Brandon last updated on 27/Sep/22

Thank you ��

Answered by som(math1967) last updated on 27/Sep/22

$$\mathrm{\angle }DAC=\mathrm{\angle }DCA=\frac{\pi }{4}$$$$\therefore \mathrm{\angle }CAB=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$$$BD=\sqrt{5^2+{12}^2}=13$$$$\therefore \mathrm{\angle }ABD={\sin }^{-1}\frac{5}{13}$$$$a=\pi -\frac{\pi }{4}-{\sin }^{-1}\frac{5}{13}$$$$sina=sin(\frac{3\pi }{4}-{\sin }^{-1}\frac{5}{13})$$$$=sin\frac{3\pi }{4}cos\left({\sin }^{-1}\frac{5}{13}\right)-cos\frac{3\pi }{4}sin{\sin }^{-1}\frac{5}{13}$$$$=\frac{1}{\sqrt{2}}\times \frac{12}{13}+\frac{1}{\sqrt{2}}\times \frac{5}{13}★$$$$=\frac{17}{13\sqrt{2}}=\frac{17\sqrt{2}}{26}$$$$★cos\left({\sin }^{-1}\frac{5}{13}\right)=cos\left({\cos }^{-1}\frac{12}{13}\right)$$

Commented by Ar Brandon last updated on 27/Sep/22

Great! Thanks!

Commented by Tawa11 last updated on 28/Sep/22

$$\mathrm{Great}\mathrm{sirs}.$$

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