a.

\(P=cos120^0+cos120^0+cos120^0=-\dfrac{3}{2}\)

b.

\(A=\dfrac{\dfrac{sinx}{cosx}-\dfrac{cosx}{cosx}}{\dfrac{sinx}{cosx}+\dfrac{cosx}{cosx}}=\dfrac{tanx-1}{tanx+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)

c.

\(A=\dfrac{cos\left(720+30\right)+sin\left(360+60\right)}{sin\left(-360+30\right)-cos\left(-360-30\right)}=\dfrac{cos30+sin60}{sin30-cos30}=-3-\sqrt{3}\)

Ta có: \(BC = \frac{{AB}}{{\cos {{30}^o}}} = 3:\frac{{\sqrt 3 }}{2} = 2\sqrt 3 \); \(AC = BC.\sin \widehat {ABC} = 2\sqrt 3 .\sin {30^o} = \sqrt 3 .\)

\(\overrightarrow {BA} .\overrightarrow {BC}  = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|\cos (\overrightarrow {BA} ,\overrightarrow {BC} ) = 3.2\sqrt 3 .\cos \widehat {ABC} = 6\sqrt 3 .\cos {30^o} = 6\sqrt 3 .\frac{{\sqrt 3 }}{2} = 9.\)

\(\overrightarrow {CA} .\overrightarrow {CB}  = \left| {\overrightarrow {CA} } \right|.\left| {\overrightarrow {CB} } \right|\cos (\overrightarrow {CA} ,\overrightarrow {CB} ) = \sqrt 3 .2\sqrt 3 .\cos \widehat {ACB} = 6.\cos {60^o} = 6.\frac{1}{2} = 3.\)

Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)

\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)

\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:

a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)

\(=\cos150^o+\sin30^o+\tan60^o\)

\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)

\(=\frac{\sqrt{3}+1}{2}\)

b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)

\(=\sin90^o+\cos30^o+\cos0^o\)

\(=1+\frac{\sqrt{3}}{2}\)

\(=\frac{2+\sqrt{3}}{2}\)

1.

\(\overrightarrow{AB}.\overrightarrow{BC}=\overrightarrow{AB}.\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}.\left(-\overrightarrow{AB}\right)+\overrightarrow{AB}.\overrightarrow{AC}=-AB^2=-25\)

2.

\(\overrightarrow{AB}.\overrightarrow{BD}=\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AD}\right)=-\overrightarrow{AB}.\overrightarrow{AB}+\overrightarrow{AB}.\overrightarrow{AD}=-AB^2+0=-64\)

Ta có: \(\overrightarrow {AB} .\overrightarrow {AC}  = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)\)

Mà \(\left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = \widehat {BAC}\)\( \Rightarrow \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = \cos \widehat {BAC}\)

Lại có: \(\cos \widehat {BAC} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)(suy ra từ định lí cosin)

\(\begin{array}{l} \Rightarrow \overrightarrow {AB} .\overrightarrow {AC}  = AB.AC.\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\ \Leftrightarrow \overrightarrow {AB} .\overrightarrow {AC}  = c.b.\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\ \Leftrightarrow \overrightarrow {AB} .\overrightarrow {AC}  = \frac{{{b^2} + {c^2} - {a^2}}}{2}\end{array}\)

\(T=\overrightarrow{GA}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\overrightarrow{GB}.\overrightarrow{CA}+\overrightarrow{GC}.\overrightarrow{AB}\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}-\overrightarrow{GA}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}-\overrightarrow{GB}\right)\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}+\overrightarrow{AG}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}+\overrightarrow{BG}\right)\)

\(=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{AC}.\overrightarrow{BA}\)

\(=0\)

a) Có \(\overrightarrow{BC}^2=\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{AC}^2+\overrightarrow{AB}^2-2\overrightarrow{AC}.\overrightarrow{AB}\)
Suy ra: \(\overrightarrow{AC}.\overrightarrow{AB}=\dfrac{\overrightarrow{AC^2}+\overrightarrow{AB}^2-\overrightarrow{BC}^2}{2}=\dfrac{8^2+6^2-11^2}{2}=-\dfrac{21}{2}\).
Do \(\overrightarrow{AC}.\overrightarrow{AB}< 0\) nên \(cos\widehat{BAC}< 0\) suy ra góc A là góc tù.
b) Từ câu a suy ra: \(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=-\dfrac{21}{2.6.8}=-\dfrac{7}{32}\).
Do N là trung điểm của AC nên \(AN=AC:2=8:2=4cm\).
\(\overrightarrow{AM}.\overrightarrow{AN}=AM.AN.cos\left(\overrightarrow{AM},\overrightarrow{AN}\right)\)
\(=2.4.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=2.4.\dfrac{-7}{32}=-\dfrac{7}{4}\).

Ta có I CA+AB I = I CB I =CB

Xét tam giác ABC ( A=90 ) áp dụng định lý pytago có

CB^2 = AB^2 + AC^2 = 9+16=25 => CB=5.

Vậy I CA+AB I= I CB I =5

\(\begin{array}{l}A{B^2} + \overrightarrow {AB} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CA}  = {\overrightarrow {AB} ^2} + \overrightarrow {AB} .\overrightarrow {BC}  + \overrightarrow {AB} .\overrightarrow {CA} \\ = \overrightarrow {AB} (\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA} ) = \overrightarrow {AB} (\overrightarrow {AC}  + \overrightarrow {CA} ) = \overrightarrow {AB} .\overrightarrow 0  = 0.\end{array}\)