Cho tam giác ABC có các góc thỏa mãn sin(A) + sin(B) = cos(A) + cos(B) . Tính số đo góc C của tam giác ABC

20/20

30⁰

90⁰

60⁰

40⁰

Giải thích

\[{\rm{sin}}\left( {\rm{A}} \right){\rm{ + sin}}\left( {\rm{B}} \right){\rm{ = cos}}\left( {\rm{A}} \right){\rm{ + cos}}\left( {\rm{B}} \right)\]

\[ \Leftrightarrow {\rm{sin}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right){\rm{ = cos}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right)\](1)

Nếu \[{\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right){\rm{ = 0}} \Rightarrow \frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}{\rm{ = 9}}{{\rm{0}}^{\rm{0}}} \Rightarrow {\rm{A}} - {\rm{B = 18}}{{\rm{0}}^{\rm{0}}}{\rm{ = A + B + C}} \Leftrightarrow {\rm{2B + C = 0}}\]

Nếu\[{\rm{cos}}\left( {\frac{{{\rm{A}} - {\rm{B}}}}{{\rm{2}}}} \right) \ne 0\] khi đó

\[\left( {\rm{1}} \right) \Leftrightarrow {\rm{sin}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{ = cos}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right) \Leftrightarrow {\rm{sin}}\left( {\frac{{{\rm{A + B}}}}{{\rm{2}}}} \right){\rm{ = sin}}\left( {\frac{{\rm{C}}}{{\rm{2}}}} \right)\]do \[\frac{{{\rm{A + B}}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{C}}}{{\rm{2}}}{\rm{ = }}{90^0}\]

\[ \Rightarrow \frac{{{\rm{A + B}}}}{{\rm{2}}}{\rm{ = }}\frac{{\rm{C}}}{{\rm{2}}} \Leftrightarrow {\rm{A + B = C}} \Leftrightarrow {\rm{18}}{{\rm{0}}^{\rm{0}}} - {\rm{C = C}} \Rightarrow {\rm{C = 9}}{{\rm{0}}^{\rm{0}}}\]

Đáp án cần chọn là: B