a) Kẻ \(DE \bot BC\)
Suy ra: \(BE = EC = \dfrac{1}{2}BC = 2,5\left( {cm} \right)\)
Trong tam giác vuông \(BDE\), ta có:
\(DE = BD.\sin \widehat {DBE}\)\( = 2,5.\sin 60^\circ = \dfrac{{5\sqrt 3 }}{2}\left( {cm} \right)\)
Trong tam giác vuông \(ADE\), ta có:
\(AD = \dfrac{{DE}}{{\sin \widehat A}} = \dfrac{{\dfrac{{5\sqrt 3 }}{2}}}{{\sin 40^\circ }}\)\( \approx 6,736\left( {cm} \right)\)
b) Trong tam giác vuông \(ADE\), ta có:
\(AE = AD.\cot g\widehat A \)\(\approx 6,736.\cos40^\circ = 5,16\left( {cm} \right)\)
Ta có: \(AB = AE - BE\)\( = 5,16 - 2,5 = 2,66\left( {cm} \right)\)
