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Question-169092

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Question Number 169092 by Shrinava last updated on 24/Apr/22
Answered by mr W last updated on 24/Apr/22
Commented by mr W last updated on 24/Apr/22
BE^2 =a^2 +b^2 −2ab cos (π−(π/2)−θ) BE^2 =a^2 +b^2 −2ab sin θ=((√2)c)^2 ...(i) replace θ with −θ we get BD^2 =a^2 +b^2 +2ab sin θ=((√2)d)^2 ...(ii) (i)+(ii): 2(a^2 +b^2 )=2c^2 +2d^2 ⇒a^2 +b^2 =c^2 +d^2 proved✓ it can also be proved that the blue square and the green square always touch each other as shown. sin (α+(π/4))=((a cos θ)/( (√(a^2 +b^2 −2ab sin θ)))) sin (β+(π/4))=((a cos θ)/( (√(a^2 +b^2 +2ab sin θ)))) ((√2)a)^2 =c^2 +d^2 +2cd sin (α+β) ? 2a^2 =c^2 +d^2 −2cd cos (α+(π/4)+β+(π/4)) b^2 −a^2 sin^2 θ=(√((a^2 sin^2 θ+b^2 )^2 −4a^2 b^2 sin^2 θ)) b^4 +a^4 sin^4 θ=(a^2 sin^2 θ+b^2 )^2 −2a^2 b^2 sin^2 θ 0=0 ✓
$${BE}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${BE}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{sin}\:\theta=\left(\sqrt{\mathrm{2}}{c}\right)^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$${replace}\:\theta\:{with}\:−\theta\:{we}\:{get} \\ $$$${BD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{sin}\:\theta=\left(\sqrt{\mathrm{2}}{d}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{2}{c}^{\mathrm{2}} +\mathrm{2}{d}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:\:\:{proved}\checkmark \\ $$$$ \\ $$$${it}\:{can}\:{also}\:{be}\:{proved}\:{that}\:{the}\:{blue} \\ $$$${square}\:{and}\:{the}\:{green}\:{square}\:{always} \\ $$$${touch}\:{each}\:{other}\:{as}\:{shown}. \\ $$$$\mathrm{sin}\:\left(\alpha+\frac{\pi}{\mathrm{4}}\right)=\frac{{a}\:\mathrm{cos}\:\theta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{sin}\:\theta}} \\ $$$$\mathrm{sin}\:\left(\beta+\frac{\pi}{\mathrm{4}}\right)=\frac{{a}\:\mathrm{cos}\:\theta}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{sin}\:\theta}} \\ $$$$\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{cd}\:\mathrm{sin}\:\left(\alpha+\beta\right)\:\:? \\ $$$$\mathrm{2}{a}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{cd}\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{4}}+\beta+\frac{\pi}{\mathrm{4}}\right) \\ $$$${b}^{\mathrm{2}} −{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta=\sqrt{\left({a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$${b}^{\mathrm{4}} +{a}^{\mathrm{4}} \mathrm{sin}^{\mathrm{4}} \:\theta=\left({a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta+{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\mathrm{0}=\mathrm{0}\:\checkmark \\ $$
Commented by Tawa11 last updated on 24/Apr/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Shrinava last updated on 24/Apr/22
Perfect dear sir thank you
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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