Question-170323

Question Number 170323 by cortano1 last updated on 20/May/22

Answered by mr W last updated on 20/May/22

((AC)/(sin 18°))=((AD)/(sin 12°)) ...(i) ((sin (α+18°))/(DB))=((sin α)/(AD)) ...(ii) (i)×(ii): ((sin (α+18°))/(sin 18°))=((sin α)/(sin 12°)) ((sin α cos 18°+cos α sin 18°)/(sin 18°))=((sin α)/(sin 12°)) (1/(tan 18°))+(1/(tan α))=(1/(sin 12°)) tan α=(1/((1/(sin 12°))−(1/(tan 18°))))=(1/( (√3))) ⇒α=30°

$$\frac{{AC}}{\mathrm{sin}\:\mathrm{18}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{12}°}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{{DB}}=\frac{\mathrm{sin}\:\alpha}{{AD}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\frac{\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{18}°+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{18}°}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{18}°}+\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{12}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{18}°}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\alpha=\mathrm{30}° \\ $$

Commented by learner22 last updated on 21/May/22