Tam giác ABC có cạnh là BC = 14, CA = 18, AB = 20, ta cần tìm các góc \(\widehat A,\widehat B,\widehat C\)
Ta có \(\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)\( = \dfrac{{{{18}^2} + {{20}^2} - {{14}^2}}}{{2.18.20}} \approx 0,7333\)
\( \Rightarrow \widehat A \approx {42^0}50'\)
\(\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\)\( = \dfrac{{{{14}^2} + {{20}^2} - {{18}^2}}}{{2.14.20}} \approx 0,4857\) \( \Rightarrow \widehat B \approx {60^0}56'\)
\(\widehat C = {180^0} - (\widehat A + \widehat B) \approx {76^0}14'\)
