Question

 giúp mình câu b với ạ

Answer 1

a) Xét tam giác \(ABD\) có :

\(\widehat{DAB}=90^o\left(HCN\right)\)

\(AD=BC=a\)

\(AB=a\sqrt{3}\)

\(\Rightarrow\Delta ABD\) là tam giác nửa đều

\(\Rightarrow\widehat{ABD}=30^o\)

b) Xét tam giác vuông \(DCE\)

\(AE=\dfrac{2}{3}AD\Rightarrow ED=\dfrac{1}{3}AD=\dfrac{a}{3}\)

\(CD=AB=a\sqrt{3}\left(HCN\right)\)

\(CE^2=ED^2+CD^2=\dfrac{a^2}{9}+3a^2=\dfrac{28a^2}{9}\left(Pitago\right)\)

\(\Rightarrow CE=\dfrac{2a\sqrt{7}}{3}\)

\(sin\widehat{ECD}=cos\widehat{DEC}=\dfrac{ED}{CE}=\dfrac{\dfrac{a}{3}}{\dfrac{2a\sqrt{7}}{3}}=\dfrac{1}{2\sqrt{7}}\)

\(cos\widehat{ECD}=sin\widehat{DEC}=\dfrac{CD}{CE}=\dfrac{a\sqrt{3}}{\dfrac{2a\sqrt{7}}{3}}=\dfrac{3\sqrt{3}}{2\sqrt{7}}\)

\(tan\widehat{ECD}=cot\widehat{DEC}=\dfrac{sin\widehat{ECD}}{cos\widehat{ECD}}=\dfrac{\dfrac{1}{2\sqrt{7}}}{\dfrac{3\sqrt{3}}{2\sqrt{7}}}=\dfrac{1}{3\sqrt{3}}\)

\(cot\widehat{ECD}=tan\widehat{DEC}=\dfrac{1}{tan\widehat{ECD}}=3\sqrt{3}\)