Áp dụng định lý $\cos$ ta được:
$AC^2 = AB^2 + BC^2 - 2AB.BC.\cos B$
$\to AC^2 = 3^2 + 4,5^2 - 2.3.4,5.\cos60^o$
$\to AC^2 = 15,75$
$\to AC = \sqrt{15,75}\,cm$
$AB^2 = AC^2 + BC^2 - 2.AC.BC.\cos C$
$\to \cos C =\dfrac{AC^2 + BC^2 - AB^2}{2AC.BC} =\dfrac{15,75 + 4,5^2 - 3^2}{2.\sqrt{15,75}.4,5}$
$\to \cos C \approx 0,07559$
$\to \widehat{C}\approx 86^o$
$\to \widehat{A}=180^o - \widehat{B}-\widehat{C}=180^o - 60^o - 86^o$
$\to \widehat{A}=34^o$