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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

side of triangle=a ∠ADE=x⇒∠FDC=30−x h=acos x (1/2)a^2 cos xsin x=5 (1/2)a^2 cos (30−x)sin (30−x)=3 ((sin 2x)/(sin (60−2x)))=(5/3) 3sin 2x=5(sin 60cos 2x−cos 60sin 2x) 6sin 2x=5(√3)cos 2x−5sin 2x tan 2x=((5(√3))/(11)) cos 2x=((11)/( (√(196))))=((11)/(14)) sin 2x=((5(√3))/(14)) S=(1/2)a^2 sin (30+x)cos (30+x) =(a^2 /4)sin (60+2x) =(a^2 /4)(sin 60cos 2x+cos 60sin 2x) =(a^2 /4)(((√3)/2)×((11)/(14))+(1/2)×((5(√3))/(14)))=((a^2 (√3))/7) (a^2 /4)sin 2x=5⇒a^2 =((20)/(sin 2x))=((20×14)/(5(√3))) S=((20×14)/(5(√3)))×((√3)/7)=8 There should be an elegent solution without requiring trigonometry. MirW, ajfour or MJS sir might suggest.

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

Answered by ajfour last updated on 06/Nov/20

Commented by ajfour last updated on 06/Nov/20

xcos (θ−30°)sin (θ−30°)=((10)/x) ⇒ x^2 sin (2θ−60°)=20 ....(i) similarly x^2 sin (120°−2θ)=12 ...(ii) also xcos θsin θ=((2S)/x) ⇒ x^2 sin 2θ=4S ....(iii) Adding (i)&(ii) 2x^2 sin 30°cos (2θ−90°)=32 ⇒ x^2 sin 2θ = 32 = 4S ⇒ S= 8 sq. units

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

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