Menu Close

Question-187893

Tinku Tara 0 Comments
Pin




Question Number 187893 by Rupesh123 last updated on 23/Feb/23
Answered by a.lgnaoui last updated on 23/Feb/23
△ABD ((sin 53)/(AD))=((sin (14+x))/(AB))=((sin (14+53+x))/(2k)) 2ksin 53=ADsin (67+x) (1) △ACD ((sin 14)/k)=((sin x)/(AD)) ksin x=ADsin 14 (2) (((1))/((2)))⇒ ((2sin 53)/(sin x))=((sin (67+x))/(sin 14)) 2sin 53×sin 14=sin x×sin (67+x) =sin^2 x×cos 67+sin xcos xsin 67 =cos 67sin^2 x+sin 67sin x(√(1−sin^2 x )) 1−sin^2 x=(2sin 53sin 14−cos 67sin^2 x)^2 ×(1/(sin^2 67sin^2 x)) posins sin x=z sin 53=a sin 14=b cos 67=c sin 67=d d^2 z^2 (1−z^2 )=(2ab−cz^2 )^2 d^2 z^4 −2d^2 z^2 +d^2 z^2 =4a^2 b^2 +c^2 z^4 −4abcz^2 z^2 =t (d^2 −c^2 )t^2 +(4abc−d^2 )t−4a^2 b^2 =0 t^2 +((4abc−d^2 )/(d^2 −c^2 ))t−((4a^2 b^2 )/(d^2 −c^2 ))=0 (t+((4abc−d^2 )/(2(d^2 −c^2 ))))^2 −(((4a^2 b^2 (d^2 −c^2 ))/(4(d^2 −c^2 )^2 ))+(((4abc−d^2 ))/(4(d^2 −c^2 )^2 )))=0 { ((t+((4abc−d^2 )/(2(d^2 −c^2 )))±((√(4a^2 b^2 (d^2 −c^2 )+4abc−d^2 ))/(2(d^2 −c^2 )))=0)),((t=sin^2 x x=arcsin (√(t.)) )) :} (a suivre)........
$$\bigtriangleup{ABD}\:\:\:\:\frac{\mathrm{sin}\:\mathrm{53}}{{AD}}=\frac{\mathrm{sin}\:\left(\mathrm{14}+{x}\right)}{{AB}}=\frac{\mathrm{sin}\:\left(\mathrm{14}+\mathrm{53}+{x}\right)}{\mathrm{2}{k}} \\ $$$$\:\:\:\mathrm{2}{k}\mathrm{sin}\:\mathrm{53}={AD}\mathrm{sin}\:\left(\mathrm{67}+{x}\right)\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ACD}\:\:\:\frac{\mathrm{sin}\:\mathrm{14}}{{k}}=\frac{\mathrm{sin}\:{x}}{{AD}} \\ $$$$\:\:\:\:\:{k}\mathrm{sin}\:{x}={AD}\mathrm{sin}\:\mathrm{14}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\frac{\left(\mathrm{1}\right)}{\left(\mathrm{2}\right)}\Rightarrow\:\:\:\frac{\mathrm{2sin}\:\mathrm{53}}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\left(\mathrm{67}+{x}\right)}{\mathrm{sin}\:\mathrm{14}} \\ $$$$\:\:\:\mathrm{2sin}\:\mathrm{53}×\mathrm{sin}\:\mathrm{14}=\mathrm{sin}\:{x}×\mathrm{sin}\:\left(\mathrm{67}+{x}\right) \\ $$$$=\mathrm{sin}\:^{\mathrm{2}} {x}×\mathrm{cos}\:\mathrm{67}+\mathrm{sin}\:{x}\mathrm{cos}\:{x}\mathrm{sin}\:\mathrm{67} \\ $$$$=\mathrm{cos}\:\mathrm{67sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:\mathrm{67sin}\:{x}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\:}\: \\ $$$$\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}=\left(\mathrm{2sin}\:\mathrm{53sin}\:\mathrm{14}−\mathrm{cos}\:\:\mathrm{67sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{67sin}\:^{\mathrm{2}} {x}} \\ $$$${posins}\:\:\:\mathrm{sin}\:{x}={z} \\ $$$$\mathrm{sin}\:\mathrm{53}={a}\:\:\:\:\mathrm{sin}\:\mathrm{14}={b}\:\:\:\mathrm{cos}\:\mathrm{67}={c} \\ $$$$\mathrm{sin}\:\mathrm{67}={d} \\ $$$${d}^{\mathrm{2}} {z}^{\mathrm{2}} \left(\mathrm{1}−{z}^{\mathrm{2}} \right)=\left(\mathrm{2}{ab}−{cz}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${d}^{\mathrm{2}} {z}^{\mathrm{4}} −\mathrm{2}{d}^{\mathrm{2}} {z}^{\mathrm{2}} +{d}^{\mathrm{2}} {z}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{c}^{\mathrm{2}} {z}^{\mathrm{4}} −\mathrm{4}{abcz}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} ={t} \\ $$$$ \\ $$$$\left({d}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\left(\mathrm{4}{abc}−{d}^{\mathrm{2}} \right){t}−\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{{t}}^{\mathrm{2}} +\frac{\mathrm{4}\boldsymbol{{abc}}−\boldsymbol{{d}}^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} }\boldsymbol{{t}}−\frac{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left(\boldsymbol{{t}}+\frac{\mathrm{4}\boldsymbol{{abc}}−\boldsymbol{{d}}^{\mathrm{2}} }{\mathrm{2}\left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)}\right)^{\mathrm{2}} −\left(\frac{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)}{\mathrm{4}\left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\left(\mathrm{4}\boldsymbol{{abc}}−\boldsymbol{{d}}^{\mathrm{2}} \right)}{\mathrm{4}\left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\begin{cases}{\boldsymbol{{t}}+\frac{\mathrm{4}\boldsymbol{{abc}}−\boldsymbol{{d}}^{\mathrm{2}} }{\mathrm{2}\left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)}\pm\frac{\sqrt{\mathrm{4}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)+\mathrm{4}\boldsymbol{{abc}}−\boldsymbol{{d}}^{\mathrm{2}} }}{\mathrm{2}\left(\boldsymbol{{d}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)}=\mathrm{0}}\\{\boldsymbol{{t}}=\mathrm{sin}^{\mathrm{2}} \:{x}\:\:\:\:\:\:\boldsymbol{{x}}=\boldsymbol{{arc}}\mathrm{sin}\:\sqrt{\boldsymbol{{t}}.}\:}\end{cases} \\ $$$$\left({a}\:{suivre}\right)…….. \\ $$
Answered by mr W last updated on 23/Feb/23
m=“median” (m/k)=((sin x)/(sin 14°)) ...(i) ((2k)/m)=((sin (53°+14°+x))/(sin 53°)) ...(ii) (i)×(ii): 2=((sin x)/(sin 14°))×((sin (53°+14°+x))/(sin 53°)) 2 sin 14° sin 53°=sin^2 x cos 67°+sin x cos x sin 67° 4 sin 14° sin 53°=(1−cos 2x) cos 67°+sin 2x sin 67° cos 67°−4 sin 14° sin 53°=cos (2x+67°) x=((cos^(−1) (cos 67°−4 sin 14° sin 53°)−67°)/2) ≈22.732°
$${m}=“{median}'' \\ $$$$\frac{{m}}{{k}}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{14}°}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{2}{k}}{{m}}=\frac{\mathrm{sin}\:\left(\mathrm{53}°+\mathrm{14}°+{x}\right)}{\mathrm{sin}\:\mathrm{53}°}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{2}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{14}°}×\frac{\mathrm{sin}\:\left(\mathrm{53}°+\mathrm{14}°+{x}\right)}{\mathrm{sin}\:\mathrm{53}°} \\ $$$$\mathrm{2}\:\mathrm{sin}\:\mathrm{14}°\:\mathrm{sin}\:\mathrm{53}°=\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}\:\mathrm{67}°+\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{67}° \\ $$$$\mathrm{4}\:\mathrm{sin}\:\mathrm{14}°\:\mathrm{sin}\:\mathrm{53}°=\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\:\mathrm{cos}\:\mathrm{67}°+\mathrm{sin}\:\mathrm{2}{x}\:\:\mathrm{sin}\:\mathrm{67}° \\ $$$$\mathrm{cos}\:\mathrm{67}°−\mathrm{4}\:\mathrm{sin}\:\mathrm{14}°\:\mathrm{sin}\:\mathrm{53}°=\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{67}°\right) \\ $$$${x}=\frac{\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{67}°−\mathrm{4}\:\mathrm{sin}\:\mathrm{14}°\:\mathrm{sin}\:\mathrm{53}°\right)−\mathrm{67}°}{\mathrm{2}} \\ $$$$\:\:\:\approx\mathrm{22}.\mathrm{732}° \\ $$
Commented by Rupesh123 last updated on 23/Feb/23
Very nice solution!

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *