Question Number 204628 by MM42 last updated on 23/Feb/24
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Answered by MM42 last updated on 23/Feb/24
Answered by A5T last updated on 23/Feb/24
Commented by A5T last updated on 23/Feb/24
$$\frac{{sin}\:\mathrm{60}}{{EB}}=\frac{{sin}\mathrm{40}}{{BC}}\Rightarrow{EB}=\frac{{BCsin}\mathrm{60}}{{sin}\mathrm{40}}=\frac{{BC}\sqrt{\mathrm{3}}}{\mathrm{2}{sin}\mathrm{40}} \\ $$$$\Rightarrow{EC}^{\mathrm{2}} =\frac{\mathrm{3}{BC}^{\mathrm{2}} }{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{40}}+{BC}^{\mathrm{2}} −\frac{{BC}^{\mathrm{2}} {cos}\mathrm{80}\sqrt{\mathrm{3}}}{{sin}\mathrm{40}} \\ $$$$\Rightarrow{EC}={BC}\sqrt{\frac{\mathrm{3}}{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{40}}−\frac{\sqrt{\mathrm{3}}{cos}\mathrm{80}}{{sin}\mathrm{40}}+\mathrm{1}} \\ $$$$\frac{{DC}}{{sin}\:\mathrm{70}}=\frac{{BC}}{{sin}\mathrm{30}}\Rightarrow{DC}=\mathrm{2}{BCsin}\mathrm{70} \\ $$$$\frac{{sin}\left(\mathrm{30}+{a}\right)}{{EC}}=\frac{{sin}\left(\mathrm{130}−{a}\right)}{{DC}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{30}°+{a}\right)}{\:\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{40}°}−\frac{\sqrt{\mathrm{3}}{cos}\mathrm{80}}{{sin}\mathrm{40}°}}}=\frac{{sin}\left(\mathrm{130}°−{a}\right)}{\mathrm{2}{sin}\mathrm{70}°} \\ $$$$\Rightarrow{a}=\mathrm{20}° \\ $$