Bài 1.
Ta có: \(A{C^2} = \sqrt {B{C^2} - A{B^2}} \)\(\;= \sqrt {{{15}^2} - {9^2}} = 12\,\left( {cm} \right)\)
\(\eqalign{ & {\mathop{\rm sinB}\nolimits} = {{AC} \over {BC}} = {{12} \over {15}} = {4 \over 5} \cr&\Rightarrow {\mathop{\rm cosC}\nolimits} = {4 \over 5} \cr & {\mathop{\rm cosB}\nolimits} = {{AB} \over {BC}} = {9 \over {15}} = {3 \over 5}\cr& \Rightarrow \sin C = {3 \over 5} \cr & \tan B = {{AC} \over {AB}} = {{12} \over 9} = {4 \over 3}\cr& \Rightarrow \cot C = {4 \over 3} \cr & \cot B = {{AB} \over {AC}} = {3 \over 4}\cr& \Rightarrow \tan C = {3 \over 4}. \cr} \)
Bài 2.
\(\eqalign{ & \cos 60^\circ = \sin \left( {90^\circ - 60^\circ } \right) = \sin 30^\circ \cr & \sin 65^\circ = \cos \left( {90^\circ - 65^\circ } \right) = \cos 25^\circ \cr & \cos 55^\circ 10' = \sin \left( {90^\circ - 55^\circ 10'} \right) \cr&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \sin 34^\circ 50' \cr & \tan 75^\circ = \cot \left( {90^\circ - 75^\circ } \right) = \cot 15^\circ \cr & \cot 80^\circ = \tan \left( {90^\circ - 80^\circ } \right) = \tan 10^\circ \cr} \)
