# Question-76948

Question Number 76948 by Master last updated on 01/Jan/20
Answered by mr W last updated on 02/Jan/20
$${let}\:\alpha=\angle{A} \\$$$$\angle{B}=\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}} \\$$$${AB}={AC}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}} \\$$$$\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right) \\$$$$\mathrm{1}=\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}−\mathrm{1}\right)\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\$$$${let}\:{t}=\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}} \\$$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\$$$$\Rightarrow{t}=\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\mathrm{0}.\mathrm{56984} \\$$$$\Rightarrow\alpha=\mathrm{33}.\mathrm{108}° \\$$
Commented by mr W last updated on 02/Jan/20
$${thanks}\:{sir}!\:{it}\:{was}\:{my}\:{error}. \\$$
Commented by \$@ty@m123 last updated on 02/Jan/20
$${Pl}.\:{clarify}\:\mathrm{3}{rd}\:{line}. \\$$$${I}\:{get}.. \\$$$$\frac{{AC}}{\mathrm{sin}\:{B}}=\frac{{BC}}{\mathrm{sin}\:{A}} \\$$$$\frac{{AC}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{sin}\:\alpha} \\$$$$\frac{{AC}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\alpha}{\mathrm{2}}\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}} \\$$$${AC}=\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\alpha}{\mathrm{2}}} \\$$$$\\$$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\$$