# Bài 6 trang 51 SGK Hình học 10 nâng cao

##### Hướng dẫn giải

a) Ta có

$$(\overrightarrow {AB} ,\,\overrightarrow {BC} ) = {150^0}\,\,;\,\,\,(\overrightarrow {BA} ,\,\overrightarrow {BC} ) = {30^0}\,\,;\,\,\,(\overrightarrow {AC} ,\,\overrightarrow {CB} ) = {120^0}$$

Do đó

\eqalign{ & \cos (\overrightarrow {AB} ,\,\overrightarrow {BC} ) + \sin (\overrightarrow {BA} ,\,\overrightarrow {BC} ) + \tan {{(\overrightarrow {AC} ,\,\overrightarrow {CB} )} \over 2} = \cos {150^0} + {\mathop{\rm s}\nolimits} {\rm{in3}}{{\rm{0}}^0} + \tan {60^0} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,{{ - \sqrt 3 } \over 2} + {1 \over 2} + \sqrt 3 = {{\sqrt 3 + 1} \over 2} \cr}

b) Ta có $$(\overrightarrow {CA} ,\,\overrightarrow {BA} ) = {90^0}$$ ,do đó

\eqalign{ & \sin (\overrightarrow {AB} ,\,\overrightarrow {AC} ) + \cos (\overrightarrow {BC} ,\,\overrightarrow {BA} ) + \cos (\overrightarrow {CA} ,\,\overrightarrow {BA} ) = \sin {90^0} + \cos {30^0} + \cos {90^0} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + {{\sqrt 3 } \over 2} + 0 = {{2 + \sqrt 3 } \over 2} \cr}