Question Number 179688 by Acem last updated on 01/Nov/22
Commented by HeferH last updated on 01/Nov/22
Commented by HeferH last updated on 01/Nov/22
$$\phi=\mathrm{30}° \\ $$$$ \\ $$
Commented by CElcedricjunior last updated on 01/Nov/22
$${calculons}\:\boldsymbol{\emptyset} \\ $$$${d}'\boldsymbol{{apr}}\grave {\boldsymbol{{e}}}{s}\:{le}\:{theoreme}\:{des}\:{sinus} \\ $$$$\frac{\boldsymbol{{x}}}{\boldsymbol{{sin}}\mathrm{15}}=\frac{\boldsymbol{{b}}}{\boldsymbol{{sin}}\left(\mathrm{30}\right)}=\frac{\boldsymbol{{k}}}{\boldsymbol{{sin}}\mathrm{135}}=>\frac{\boldsymbol{{b}}}{\boldsymbol{{x}}}=\mathrm{2}\boldsymbol{{cos}}\mathrm{15}\:\left(\mathrm{1}\right) \\ $$$$\frac{\boldsymbol{{x}}}{\boldsymbol{{sin}\emptyset}}=\frac{\boldsymbol{{b}}}{\boldsymbol{{sin}}\mathrm{45}}=\frac{\boldsymbol{{m}}}{\boldsymbol{{sin}\Omega}}=>\frac{\boldsymbol{{b}}}{\boldsymbol{{x}}}=\frac{\boldsymbol{{sin}}\mathrm{45}}{\boldsymbol{{sin}\emptyset}}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)=\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow\frac{\boldsymbol{{sin}}\mathrm{45}}{\boldsymbol{{sin}\emptyset}}=\mathrm{2}\boldsymbol{{cos}}\mathrm{15} \\ $$$$\Leftrightarrow\boldsymbol{{sin}\emptyset}=\frac{\boldsymbol{{sin}}\mathrm{45}}{\mathrm{2}\boldsymbol{{cos}}\mathrm{15}}=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\sqrt{\mathrm{3}}+\mathrm{2}}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}}} \\ $$$$\boldsymbol{{sin}\emptyset}=\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}}}{\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4}} \\ $$$$=>\boldsymbol{\emptyset}=\boldsymbol{{arsin}}\left(\frac{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}}\right)=\mathrm{21}.\mathrm{47}° \\ $$$$……….{le}\:{celebre}\:{cedric}\:{junior}…………. \\ $$
Commented by Acem last updated on 01/Nov/22
$${Sir}\:{HeferH},\:{the}\:{same}\:{way}\:{of}\:{mine},\:{thanks} \\ $$
Commented by Acem last updated on 01/Nov/22
$${Monsieur}\:{Junior},\:{merci}\:{beaucoup}, \\ $$$${mais}\:{l}'{angle}\:{est}\:\mathrm{30}° \\ $$
Answered by mr W last updated on 01/Nov/22
$${an}\:{other}\:{way} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{45}+\phi\right)}{\mathrm{sin}\:\phi}=\frac{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{15}\right)}{\mathrm{sin}\:\mathrm{15}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\phi}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{15}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\phi=\mathrm{30}° \\ $$
Commented by Acem last updated on 01/Nov/22
$${Good}! \\ $$