Question Number 170323 by cortano1 last updated on 20/May/22
Answered by mr W last updated on 20/May/22
$$\frac{{AC}}{\mathrm{sin}\:\mathrm{18}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{12}°}\:\:\:...\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{{DB}}=\frac{\mathrm{sin}\:\alpha}{{AD}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\frac{\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{18}°+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{18}°}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{18}°}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\alpha=\mathrm{30}° \\ $$
Commented by learner22 last updated on 21/May/22
$$\boldsymbol{{Nice}} \\ $$
Commented by cortano1 last updated on 22/May/22
$${how}\:{to}\:{show}\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{18}°}=\sqrt{\mathrm{3}}\:? \\ $$
Commented by Tawa11 last updated on 08/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$