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Question-165170

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Question Number 165170 by Tawa11 last updated on 26/Jan/22
Commented by mr W last updated on 27/Jan/22
Commented by mr W last updated on 27/Jan/22
generally: if k+m=2n, then x=((45°)/n)
$${generally}: \\ $$$${if}\:{k}+{m}=\mathrm{2}{n},\:{then} \\ $$$${x}=\frac{\mathrm{45}°}{{n}} \\ $$
Commented by Tawa11 last updated on 27/Jan/22
I appreciate sir.
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Answered by aleks041103 last updated on 27/Jan/22
Commented by aleks041103 last updated on 27/Jan/22
obvious constraint: 7x<90° ⇒x<12(6/7)° ((OC)/(OB))=tg(5x) ((OB)/(OA))=ctg(3x) ((OA)/(OD))=tg(5x) ((OD)/(OC))=ctg(7x) ⇒((tg^2 (5x))/(tg(3x)tg(7x)))=1 ⇒tg(3x)tg(7x)=tg^2 (5x) tg(10x)=((2tg(5x))/(1−tg^2 (5x)))=((tg(3x)+tg(7x))/(1−tg(3x)tg(7x))) ⇒2tg(5x)=tg(3x)+tg(7x) or tg(3x)tg(7x)=tg^2 (5x)=1 1)2tg(5x)=tg(3x)+tg(7x) ⇒2tg(5x)=((sin(10x))/(cos(3x)cos(7x)))=((2sin(5x)cos(5x))/(cos(3x)cos(7x)))=((2sin(5x))/(cos(5x))) ⇒cos^2 (5x)=cos(3x)cos(7x) but tg(3x)tg(7x)=tg^2 (5x) ⇒sin^2 (5x)=sin(3x)sin(7x) ⇒1=cos^2 (5x)+sin^2 (5x)= =cos(3x)cos(7x)+sin(3x)sin(7x)= =cos(4x)=1 ⇒4x=k360° ⇒x=k.90° but x<13°⇒k=0⇒x=0 →trivial 2)tg^2 (5x)=tg(3x)tg(7x)=1 ⇒tg(5x)=±1 5x=45°,135°,225°,315° ⇒x=9°,27°,45°,63° but x<13°⇒x=9° only possible check: tg(27°)tg(63°)=tg(27°)ctg(90−63)= =tg(27°)ctg(27°)=1✓ ⇒x=9°=(π/(20)) rad
$${obvious}\:{constraint}: \\ $$$$\mathrm{7}{x}<\mathrm{90}° \\ $$$$\Rightarrow{x}<\mathrm{12}\frac{\mathrm{6}}{\mathrm{7}}° \\ $$$$\frac{{OC}}{{OB}}={tg}\left(\mathrm{5}{x}\right) \\ $$$$\frac{{OB}}{{OA}}={ctg}\left(\mathrm{3}{x}\right) \\ $$$$\frac{{OA}}{{OD}}={tg}\left(\mathrm{5}{x}\right) \\ $$$$\frac{{OD}}{{OC}}={ctg}\left(\mathrm{7}{x}\right) \\ $$$$\Rightarrow\frac{{tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right)}{{tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)}=\mathrm{1} \\ $$$$\Rightarrow{tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)={tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right) \\ $$$${tg}\left(\mathrm{10}{x}\right)=\frac{\mathrm{2}{tg}\left(\mathrm{5}{x}\right)}{\mathrm{1}−{tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right)}=\frac{{tg}\left(\mathrm{3}{x}\right)+{tg}\left(\mathrm{7}{x}\right)}{\mathrm{1}−{tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)} \\ $$$$\Rightarrow\mathrm{2}{tg}\left(\mathrm{5}{x}\right)={tg}\left(\mathrm{3}{x}\right)+{tg}\left(\mathrm{7}{x}\right)\:{or}\:{tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)={tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right)=\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\mathrm{2}{tg}\left(\mathrm{5}{x}\right)={tg}\left(\mathrm{3}{x}\right)+{tg}\left(\mathrm{7}{x}\right) \\ $$$$\Rightarrow\mathrm{2}{tg}\left(\mathrm{5}{x}\right)=\frac{{sin}\left(\mathrm{10}{x}\right)}{{cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{7}{x}\right)}=\frac{\mathrm{2}{sin}\left(\mathrm{5}{x}\right){cos}\left(\mathrm{5}{x}\right)}{{cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{7}{x}\right)}=\frac{\mathrm{2}{sin}\left(\mathrm{5}{x}\right)}{{cos}\left(\mathrm{5}{x}\right)} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} \left(\mathrm{5}{x}\right)={cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{7}{x}\right) \\ $$$${but}\:{tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)={tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \left(\mathrm{5}{x}\right)={sin}\left(\mathrm{3}{x}\right){sin}\left(\mathrm{7}{x}\right) \\ $$$$\Rightarrow\mathrm{1}={cos}^{\mathrm{2}} \left(\mathrm{5}{x}\right)+{sin}^{\mathrm{2}} \left(\mathrm{5}{x}\right)= \\ $$$$={cos}\left(\mathrm{3}{x}\right){cos}\left(\mathrm{7}{x}\right)+{sin}\left(\mathrm{3}{x}\right){sin}\left(\mathrm{7}{x}\right)= \\ $$$$={cos}\left(\mathrm{4}{x}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{x}={k}\mathrm{360}° \\ $$$$\Rightarrow{x}={k}.\mathrm{90}° \\ $$$${but}\:{x}<\mathrm{13}°\Rightarrow{k}=\mathrm{0}\Rightarrow{x}=\mathrm{0}\:\rightarrow{trivial} \\ $$$$\left.\mathrm{2}\right){tg}^{\mathrm{2}} \left(\mathrm{5}{x}\right)={tg}\left(\mathrm{3}{x}\right){tg}\left(\mathrm{7}{x}\right)=\mathrm{1} \\ $$$$\Rightarrow{tg}\left(\mathrm{5}{x}\right)=\pm\mathrm{1} \\ $$$$\mathrm{5}{x}=\mathrm{45}°,\mathrm{135}°,\mathrm{225}°,\mathrm{315}° \\ $$$$\Rightarrow{x}=\mathrm{9}°,\mathrm{27}°,\mathrm{45}°,\mathrm{63}° \\ $$$${but}\:{x}<\mathrm{13}°\Rightarrow{x}=\mathrm{9}°\:{only}\:{possible} \\ $$$${check}: \\ $$$${tg}\left(\mathrm{27}°\right){tg}\left(\mathrm{63}°\right)={tg}\left(\mathrm{27}°\right){ctg}\left(\mathrm{90}−\mathrm{63}\right)= \\ $$$$={tg}\left(\mathrm{27}°\right){ctg}\left(\mathrm{27}°\right)=\mathrm{1}\checkmark \\ $$$$ \\ $$$$\Rightarrow{x}=\mathrm{9}°=\frac{\pi}{\mathrm{20}}\:{rad} \\ $$
Commented by Tawa11 last updated on 27/Jan/22
I appreciate sir
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$
Answered by nurtani last updated on 27/Jan/22
7x+5x+5x+3x = 180° 20x = 180° x = ((180°)/(20)) = 9°
$$\mathrm{7}{x}+\mathrm{5}{x}+\mathrm{5}{x}+\mathrm{3}{x}\:=\:\mathrm{180}° \\ $$$$\mathrm{20}{x}\:=\:\mathrm{180}° \\ $$$$\:\:\:\:\:{x}\:=\:\frac{\mathrm{180}°}{\mathrm{20}}\:=\:\mathrm{9}° \\ $$
Commented by mr W last updated on 27/Jan/22
i think it′s just by accident true. the method can′t be generally applied, for example in following case:
$${i}\:{think}\:{it}'{s}\:{just}\:{by}\:{accident}\:{true}.\: \\ $$$${the}\:{method}\:{can}'{t}\:{be}\:{generally}\:{applied}, \\ $$$${for}\:{example}\:{in}\:{following}\:{case}: \\ $$
Commented by mr W last updated on 27/Jan/22

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