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Question Number 72934 by Maclaurin Stickker last updated on 05/Nov/19

Answered by mr W last updated on 05/Nov/19

Commented by mr W last updated on 05/Nov/19

β+γ=45° CE=((CA)/(cos γ))=(b/(cos γ))=(b/(cos (45−β))) BD=((AB)/(cos β))=(c/(cos β)) k=((CE×BD)/(BC^2 ))=(1/(cos (45−β)cos β ))×((bc)/a^2 ) (b/a)=sin 2β (c/a)=cos 2β k=((sin 2β cos 2β)/(cos (45−β)cos β )) k=((2(√2) sin β (cos^2 β−sin^2 β))/(cos β+sin β )) k=2(√2) sin β (cos β−sin β) k=(√2) (sin 2β−1+cos 2β) (k/(√2))+1=sin 2β+cos 2β (k/(√2))+1=sin B+cos B (k/(√2))+1=(√2) sin (B+45) ((k+(√2))/2)=sin (B+45) ⇒∠B=sin^(−1) ((k+(√2))/2)−45° 0<∠B<90° ((√2)/2)<((k+(√2))/2)≤1 (√2)<k+(√2)≤2 ⇒0<k≤2−(√2)

$$\beta+\gamma=\mathrm{45}° \\ $$$${CE}=\frac{{CA}}{\mathrm{cos}\:\gamma}=\frac{{b}}{\mathrm{cos}\:\gamma}=\frac{{b}}{\mathrm{cos}\:\left(\mathrm{45}−\beta\right)} \\ $$$${BD}=\frac{{AB}}{\mathrm{cos}\:\beta}=\frac{{c}}{\mathrm{cos}\:\beta} \\ $$$${k}=\frac{{CE}×{BD}}{{BC}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{45}−\beta\right)\mathrm{cos}\:\beta\:}×\frac{{bc}}{{a}^{\mathrm{2}} } \\ $$$$\frac{{b}}{{a}}=\mathrm{sin}\:\mathrm{2}\beta \\ $$$$\frac{{c}}{{a}}=\mathrm{cos}\:\mathrm{2}\beta \\ $$$${k}=\frac{\mathrm{sin}\:\mathrm{2}\beta\:\mathrm{cos}\:\mathrm{2}\beta}{\mathrm{cos}\:\left(\mathrm{45}−\beta\right)\mathrm{cos}\:\beta\:} \\ $$$${k}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\beta\:\left(\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{sin}^{\mathrm{2}} \:\beta\right)}{\mathrm{cos}\:\beta+\mathrm{sin}\:\beta\:} \\ $$$${k}=\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\beta\:\left(\mathrm{cos}\:\beta−\mathrm{sin}\:\beta\right) \\ $$$${k}=\sqrt{\mathrm{2}}\:\left(\mathrm{sin}\:\mathrm{2}\beta−\mathrm{1}+\mathrm{cos}\:\mathrm{2}\beta\right) \\ $$$$\frac{{k}}{\sqrt{\mathrm{2}}}+\mathrm{1}=\mathrm{sin}\:\mathrm{2}\beta+\mathrm{cos}\:\mathrm{2}\beta \\ $$$$\frac{{k}}{\sqrt{\mathrm{2}}}+\mathrm{1}=\mathrm{sin}\:{B}+\mathrm{cos}\:{B} \\ $$$$\frac{{k}}{\sqrt{\mathrm{2}}}+\mathrm{1}=\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({B}+\mathrm{45}\right) \\ $$$$\frac{{k}+\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{sin}\:\left({B}+\mathrm{45}\right) \\ $$$$\Rightarrow\angle{B}=\mathrm{sin}^{−\mathrm{1}} \frac{{k}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{45}° \\ $$$$ \\ $$$$\mathrm{0}<\angle{B}<\mathrm{90}° \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}<\frac{{k}+\sqrt{\mathrm{2}}}{\mathrm{2}}\leqslant\mathrm{1} \\ $$$$\sqrt{\mathrm{2}}<{k}+\sqrt{\mathrm{2}}\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{0}<{k}\leqslant\mathrm{2}−\sqrt{\mathrm{2}} \\ $$

Commented by Maclaurin Stickker last updated on 06/Nov/19

perfect

$${perfect} \\ $$

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