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Question-40031

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Question Number 40031 by ajfour last updated on 15/Jul/18
Commented by ajfour last updated on 16/Jul/18
CE and DF ⊥ AF . Find θ.
$${CE}\:{and}\:{DF}\:\bot\:{AF}\:.\:{Find}\:\theta. \\ $$
Answered by ajfour last updated on 21/Jul/18
let A be origin and x-axis leftwards. AF=sin θ x_C =sin θ−cos θ eq. of BF : y=−((sin θ)/(sin θ+cos θ))(x−sin θ) eq. of AD : y=((cos θ)/(sin θ)) x so for x_E : −((sin θ)/(sin θ+cos θ))(x_E −sin θ)=((cos θ)/(sin θ))x_E ⇒ x_E ( ((cos θ)/(sin θ))+((sin θ)/(sin θ+cos θ)))=((sin^2 θ)/(sin θ+cos θ)) ⇒ x_E = ((sin^3 θ)/(sin θcos θ+1)) as x_E = x_C we have sin θ−cos θ = ((sin^3 θ)/(sin θcos θ+1)) ⇒ sin^2 θcos θ+sin θ−sin θcos^2 θ −cos θ=sin^3 θ ⇒ sin^2 θcos θ+sin θ−sin θcos^2 θ −cos θ=sin θ−sin θcos^2 θ ⇒ sin^2 θcos θ=cos θ ⇒ sin θ=1 or θ=(π/2) .
$${let}\:{A}\:{be}\:{origin}\:{and}\:{x}-{axis}\:{leftwards}. \\ $$$${AF}=\mathrm{sin}\:\theta \\ $$$${x}_{{C}} =\mathrm{sin}\:\theta−\mathrm{cos}\:\theta \\ $$$${eq}.\:{of}\:{BF}\::\:\:{y}=−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\left({x}−\mathrm{sin}\:\theta\right)\: \\ $$$${eq}.\:{of}\:{AD}\::\:\:{y}=\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\:{x} \\ $$$${so}\:\:{for}\:{x}_{{E}} \:: \\ $$$$−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\left({x}_{{E}} −\mathrm{sin}\:\theta\right)=\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}{x}_{{E}} \\ $$$$\Rightarrow\:{x}_{{E}} \:\left(\:\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}+\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta}\right)=\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta+\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:{x}_{{E}} =\:\frac{\mathrm{sin}\:^{\mathrm{3}} \theta}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta+\mathrm{1}} \\ $$$${as}\:\:\:{x}_{{E}} =\:{x}_{{C}} \:\:{we}\:{have} \\ $$$$\:\:\:\:\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\:=\:\frac{\mathrm{sin}\:^{\mathrm{3}} \theta}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta+\mathrm{1}} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{cos}\:\theta=\mathrm{sin}\:^{\mathrm{3}} \theta \\ $$$$\Rightarrow\:\:\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta+\mathrm{sin}\:\theta−\mathrm{sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{cos}\:\theta=\mathrm{sin}\:\theta−\mathrm{sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:\theta=\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\mathrm{sin}\:\theta=\mathrm{1}\:\:\:{or}\:\:\theta=\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by ajfour last updated on 21/Jul/18
but i had obtained some acute angle value for θ when i′d solved before; dunnow why i couldn′t obtain the same now ..
$${but}\:{i}\:{had}\:{obtained}\:{some}\:{acute} \\ $$$${angle}\:{value}\:{for}\:\theta\:{when}\:{i}'{d}\:{solved} \\ $$$${before};\:{dunnow}\:{why}\:{i}\:{couldn}'{t} \\ $$$${obtain}\:{the}\:{same}\:{now}\:.. \\ $$

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