Question Number 179938 by Acem last updated on 04/Nov/22
Commented by Acem last updated on 04/Nov/22
$${It}'{s}\:{correct},\:{Thanks}\:{Sir}! \\ $$
Commented by HeferH last updated on 04/Nov/22
Commented by HeferH last updated on 04/Nov/22
$${x}\sqrt{\mathrm{3}}\:−\:{x}\:=\:\mathrm{1} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\:=\:\frac{\sqrt{\mathrm{3}}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{tan}\:\beta\:\:=\:\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}\:−\:{x}}\:=\:\frac{\mathrm{3}\:+\:\sqrt{\mathrm{3}}}{\mathrm{3}−\:\sqrt{\mathrm{3}}}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{3}} \\ $$$$\:\beta\:=\:\mathrm{tan}^{−\mathrm{1}} \:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)\: \\ $$$$\:\beta\:=\:\mathrm{75}\:° \\ $$
Answered by mr W last updated on 04/Nov/22
$$\frac{\mathrm{2}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\mathrm{60}\right)}=\frac{\mathrm{1}×\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\mathrm{15}} \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\mathrm{60}\right)}=\frac{\mathrm{4}}{\:\mathrm{2}\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{tan}\:\beta}+\mathrm{cos}\:\mathrm{60}=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$\mathrm{tan}\:\beta=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\beta=\mathrm{75}° \\ $$
Commented by Acem last updated on 05/Nov/22
$${Very}\:{good}\:{Sir}!\:{thanks} \\ $$
Answered by Acem last updated on 05/Nov/22
Commented by Acem last updated on 05/Nov/22
$$\lambda=\:\mathrm{60}−\mathrm{45}=\:\mathrm{15} \\ $$$${MN}=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{60}=\:\mathrm{1}\:\Rightarrow\:{MN}={MA}\:\Rightarrow\:\alpha=\:\phi=\:\mathrm{30} \\ $$$$\:\Rightarrow\Upsilon=\:\lambda\:\Rightarrow\:{NA}=\:{NC} \\ $$$$\:\alpha=\:\beta_{\mathrm{1}} \:\:\:\:\:\Rightarrow\:{NA}=\:{NB}=\:{NC}\:\Rightarrow\:\theta=\:\beta_{\mathrm{2}} =\:\mathrm{45} \\ $$$$\:\Rightarrow\:\boldsymbol{\beta}=\:\mathrm{75}° \\ $$
Commented by mr W last updated on 05/Nov/22
$${nice}\:{solution}! \\ $$