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Question Number 62289 by naka3546 last updated on 19/Jun/19

Answered by MJS last updated on 19/Jun/19

$$\mathrm{we}\mathrm{have}$$$$A_1=3=\frac{1}{2}\times a\times x\Rightarrow x=\frac{6}{a}$$$$A_2=5=\frac{1}{2}\times b\times y\Rightarrow y=\frac{10}{b}$$$$\mathrm{side}\mathrm{of}\mathrm{triangle}=c$$$$\left(1\right)a^2+\frac{36}{a^2}=c^2$$$$\left(2\right)b^2+\frac{100}{b^2}=c^2$$$$\left(3\right){(a-\frac{10}{b})}^2+{(b-\frac{6}{a})}^2=c^2$$$$\mathrm{solving}\mathrm{these}\mathrm{leads}\mathrm{to}$$$$a=\sqrt[4]{972}b=\sqrt[4]{\frac{2500}{3}}c=\sqrt[4]{\frac{3136}{3}}$$$$\mathrm{or}$$$$a=\sqrt[4]{\frac{4}{3}}b=\sqrt[4]{12}c=\sqrt[4]{\frac{3136}{3}}$$$$\Rightarrow A_3=8$$

Commented by MJS last updated on 19/Jun/19

$$\mathrm{btw}\mathrm{only}\mathrm{the}{1}^{\mathrm{st}}\mathrm{solution}\mathrm{looks}\mathrm{like}\mathrm{in}\mathrm{the}$$$$\mathrm{picture}$$

Commented by mr W last updated on 19/Jun/19

$$exactsolution,verynicesir!$$$$ithink{A}_3=A_1+A_2,proofseebelow.$$

Answered by mr W last updated on 19/Jun/19

Commented by mr W last updated on 19/Jun/19

$$\phi =\frac{\pi }{2}-\frac{\pi }{3}-\theta =\frac{\pi }{6}-\theta$$$$\varphi =\pi -(\frac{\pi }{2}-\theta )-\frac{\pi }{3}=\frac{\pi }{6}+\theta$$$$A_1=\frac{c^2\sin 2\theta }{4}$$$$A_2=\frac{c^2\sin 2\phi }{4}=\frac{c^2\sin (\frac{\pi }{3}-2\theta )}{4}$$$$A_3=\frac{c^2\sin 2\varphi }{4}=\frac{c^2\sin (\frac{\pi }{3}+2\theta )}{4}$$$$\frac{A_2}{A_1}=\frac{\sin (\frac{\pi }{3}-2\theta )}{\sin 2\theta }$$$$\frac{\frac{\sqrt{3}}{2}\cos 2\theta -\frac{1}{2}\sin 2\theta }{\sin 2\theta }=\frac{A_2}{A_1}$$$$\Rightarrow \frac{\sqrt{3}}{\tan 2\theta }=\frac{2A_2}{A_1}+1$$$$\frac{A_3}{A_1}=\frac{\sin (\frac{\pi }{3}+2\theta )}{\sin 2\theta }$$$$=\frac{1}{2}(\frac{\sqrt{3}}{\tan 2\theta }+1)=\frac{1}{2}(\frac{2A_2}{A_1}+1+1)=1+\frac{A_2}{A_1}$$$$\Rightarrow A_3=A_1(1+\frac{A_2}{A_1})=A_1+A_2=3+5=8$$

Commented by Tony Lin last updated on 20/Jun/19

$$2(A_1+A_2)=c^{2}sin\theta cos\theta +c^{2}sin(\frac{\pi }{6}-\theta )cos(\frac{\pi }{6}-\theta )$$$$=\frac{1}{2}{c}^2[sin2\theta +sin(\frac{\pi }{3}-2\theta )]$$$$=\frac{1}{2}{c}^2(\frac{\sqrt{3}}{2}cos2\theta +\frac{1}{2}sin2\theta )$$$$=\frac{1}{2}{c}^{2}sin(\frac{\pi }{3}+2\theta )$$$$=c^{2}cos(\frac{\pi }{6}+\theta )sin(\frac{\pi }{6}+\theta )$$$$=2A_3$$

Answered by ajfour last updated on 19/Jun/19

$$c^2\sin \theta \cos \theta =6$$$$c^2\sin \varphi \cos \varphi =10\Rightarrow \frac{\sin 2\theta }{\sin 2\varphi }=\frac{3}{5}$$$$\theta +\varphi =\frac{\pi }{6}\Rightarrow 2\theta +2\varphi =\frac{\pi }{3}$$$$\Rightarrow 5\sin 2\theta =3\sin (\frac{\pi }{3}-2\theta )$$$$\Rightarrow 10\sin 2\theta =3\sqrt{3}\cos 2\theta -3\sin 2\theta$$$$\Rightarrow \tan 2\theta =\frac{3\sqrt{3}}{13}\Rightarrow \cos 2\theta =\frac{13}{14}$$$$\mathrm{\&}\tan 2\varphi =\tan (\frac{\pi }{3}-2\theta )$$$$=\frac{\sqrt{3}-\frac{3\sqrt{3}}{13}}{1+\frac{9}{13}}=\frac{5\sqrt{3}}{11}$$$$\Rightarrow \cos 2\varphi =\frac{11}{14}$$$$\cos 2\theta +\cos 2\varphi =2\cos (\theta +\varphi )\cos (\theta -\varphi )$$$$\Rightarrow \frac{12}{7}=\sqrt{3}\cos (\theta -\varphi )$$$$\Rightarrow \cos (\theta -\varphi )=\frac{12}{7\sqrt{3}}$$$$c^2=\frac{12}{\sin 2\theta }=\frac{12}{\left(3\sqrt{3}/14\right)}=\frac{56}{\sqrt{3}}$$$$c^2(\cos \theta -\sin \varphi )(\cos \varphi -\sin \theta )=2A_3$$$$\Rightarrow 2A_3=c^2(\cos \theta \cos \varphi +\sin \theta \sin \varphi )$$$$-c^2(\cos \theta \sin \theta +\sin \varphi \cos \varphi )$$$$\Rightarrow 2A_3=c^2\cos (\theta -\varphi )-16$$$$=\frac{56}{\sqrt{3}}\times \frac{12}{7\sqrt{3}}-16=32-16$$$$\Rightarrow {\mathit{A}}_3=8.$$

Commented by ajfour last updated on 19/Jun/19

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