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Question Number 120724 by A8;15: last updated on 02/Nov/20

Commented by A8;15: last updated on 02/Nov/20

find S

Commented by prakash jain last updated on 02/Nov/20

$$\mathrm{side}\mathrm{of}\mathrm{triangle}=\mathrm{a}$$$$\mathrm{\angle }ADE=x\Rightarrow \mathrm{\angle }FDC=30-x$$$$h=a\cos x$$$$\frac{1}{2}{a}^2\cos x\sin x=5$$$$\frac{1}{2}{a}^2\cos (30-x)\sin (30-x)=3$$$$\frac{\sin 2x}{\sin (60-2x)}=\frac{5}{3}$$$$3\sin 2x=5(\sin 60\cos 2x-\cos 60\sin 2x)$$$$6\sin 2x=5\sqrt{3}\cos 2x-5\sin 2x$$$$\tan 2x=\frac{5\sqrt{3}}{11}$$$$\cos 2x=\frac{11}{\sqrt{196}}=\frac{11}{14}$$$$\sin 2x=\frac{5\sqrt{3}}{14}$$$$S=\frac{1}{2}{a}^2\sin (30+x)\cos (30+x)$$$$=\frac{a^2}{4}\sin (60+2x)$$$$=\frac{a^2}{4}(\sin 60\cos 2x+\cos 60\sin 2x)$$$$=\frac{a^2}{4}(\frac{\sqrt{3}}{2}\times \frac{11}{14}+\frac{1}{2}\times \frac{5\sqrt{3}}{14})=\frac{a^2\sqrt{3}}{7}$$$$\frac{a^2}{4}\sin 2x=5\Rightarrow a^2=\frac{20}{\sin 2x}=\frac{20\times 14}{5\sqrt{3}}$$$$S=\frac{20\times 14}{5\sqrt{3}}\times \frac{\sqrt{3}}{7}=8$$$$\mathrm{There}\mathrm{should}\mathrm{be}\mathrm{an}\mathrm{elegent}\mathrm{solution}\mathrm{without}$$$$\mathrm{requiring}\mathrm{trigonometry}.$$$$\mathrm{M}i\mathrm{rW},\mathrm{ajfour}\mathrm{or}\mathrm{MJS}\mathrm{sir}\mathrm{might}\mathrm{suggest}.$$

Answered by ajfour last updated on 06/Nov/20

Commented by ajfour last updated on 06/Nov/20

$$x\cos (\theta -30°)\sin (\theta -30°)=\frac{10}{x}$$$$\Rightarrow x^2\sin (2\theta -60°)=20....\left(i\right)$$$$similarly$$$$x^2\sin (120°-2\theta )=12...\left(ii\right)$$$$alsox\cos \theta \sin \theta =\frac{2S}{x}$$$$\Rightarrow x^2\sin 2\theta =4S....\left(iii\right)$$$$Adding\left(i\right)\mathrm{\&}\left(ii\right)$$$$2x^2\sin 30°\cos (2\theta -90°)=32$$$$\Rightarrow x^2\sin 2\theta =32=4S$$$$\Rightarrow S=8sq.units$$

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