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Bài 35. Giải tam giác \(ABC\), biết
a) \(a = 14,\,\,b = 18,\,\,c = 20\);
b) \(a = 6,\,\,b = 7,3,\,\,c = 4,8\);
c) \(a = 4,\,\,b = 5,\,\,c = 7\)
a) Áp dụng định lí cosin ta có
\(\eqalign{
& \cos A = {{{b^2} + {c^2} – {a^2}} \over {2bc}} = {{{{18}^2} + {{20}^2} – {{14}^2}} \over {2.18.20}} \approx 0,73 \cr
& \cos B = {{{a^2} + {c^2} – {b^2}} \over {2ac}} = {{{{14}^2} + {{20}^2} – {{18}^2}} \over {2.14.20}} \approx 0,49 \cr
& \Rightarrow \,\,\,\widehat A \approx {43^0}\,\,\,\,\widehat B \approx {61^0}\,\,\,\widehat C \approx {76^0}. \cr} \)
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b) Áp dụng định lí cosin ta có
\(\eqalign{
& \cos A = {{{b^2} + {c^2} – {a^2}} \over {2bc}} = {{{{(7,3)}^2} + {{(4,8)}^2} – {6^2}} \over {2.(7,3).(4.8)}} \approx 0,58 \cr
& \cos B = {{{a^2} + {c^2} – {b^2}} \over {2ac}} = {{{6^2} + {{(4,8)}^2} – {{(7,3)}^2}} \over {2.6.(4,8)}} \approx 0,1 \cr
& \Rightarrow \,\,\,\widehat A \approx {55^0}\,\,\,\,\widehat B \approx {85^0}\,\,\,\widehat C \approx {40^0}. \cr} \)
c) Áp dụng định lí cosin ta có
\(\eqalign{
& \cos A = {{{b^2} + {c^2} – {a^2}} \over {2bc}} = {{{5^2} + {7^2} – {4^2}} \over {2.5.7}} \approx 0,83 \cr
& \cos B = {{{a^2} + {c^2} – {b^2}} \over {2ac}} = {{{4^2} + {7^2} – {5^2}} \over {2.4.7}} \approx 0,71 \cr
& \Rightarrow \,\,\,\widehat A \approx {34^0}\,\,\,\,\widehat B \approx {44^0}\,\,\,\widehat C \approx {102^0}. \cr} \)