Tính các góc chưa biết của tam giác ABC trong các trường hợp sau: AB = 15, AC = 25, BC = 30.

Theo định lí côsin ta có: AB² = BC² + AC² – 2BC.AC.cos\(\widehat {\rm{C}}\)

⇒ cos\(\widehat {\rm{C}}\) = \(\frac{{{\rm{B}}{{\rm{C}}^2} + {\rm{A}}{{\rm{C}}^2} - {\rm{A}}{{\rm{B}}^2}}}{{2.{\rm{BC}}.{\rm{AC}}}}\) = \(\frac{{{{30}^2} + {{25}^2} - {{15}^2}}}{{2.30.25}}\) = \(\frac{{13}}{{15}}\) ⇒ \(\widehat {\rm{C}}\) ≈ 29°55’.

Tương tự như trên, ta có:

cos\(\widehat {\rm{A}}\) = \(\frac{{{\rm{A}}{{\rm{B}}^2} + {\rm{A}}{{\rm{C}}^2} - {\rm{BC}}{^2}}}{{2.{\rm{AB}}.{\rm{AC}}}}\)= \(\frac{{{{15}^2} + {{25}^2} - {{30}^2}}}{{2.15.25}}\) = \(\frac{{ - 1}}{{15}}\) ⇒ \(\widehat {\rm{A}}\) ≈ 93°49’.

cos\(\widehat {\rm{B}}\) = \(\frac{{{\rm{A}}{{\rm{B}}^2} + {\rm{B}}{{\rm{C}}^2} - {\rm{AC}}{^2}}}{{2.{\rm{AB}}.{\rm{BC}}}}\)= \(\frac{{{{15}^2} + {{30}^2} - {{25}^2}}}{{2.15.30}}\) = \(\frac{5}{9}\) ⇒ \(\widehat {\rm{B}}\) ≈ 56°15’.