Định lí cosin trong tam giác

Ta có: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

$\Rightarrow \dfrac{\dfrac{b+c}{2}}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

$\Leftrightarrow \dfrac{b+c}{2\sin A}=\dfrac{b+c}{\sin B+\sin C}$

$\Leftrightarrow \sin B+\sin C=2\sin A$

Ta có: ${{AC}^{2}}={{BC}^{2}}+{{AB}^{2}}-2ac\cos B={{8}^{2}}+{{5}^{2}}-2.8.5.\cos {{60}^{0}}=49\Rightarrow AC=7$

Ta có: $\overrightarrow{AB}=(2;-2)\Rightarrow AB=2\sqrt{2}$ , $\overrightarrow{AC}=(5;1)\Rightarrow AC=\sqrt{26}$ , $\overrightarrow{BC}=(3;3)\Rightarrow BC=3\sqrt{2}$

Mặt khác $\overrightarrow{AB}.\overrightarrow{BC}=0\Rightarrow AB\bot BC$ .

Suy ra: ${{S}_{\Delta ABC}}=\dfrac{1}{2}AB.BC=6$

Ta có: $\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=\dfrac{{{13}^{2}}+{{15}^{2}}-{{14}^{2}}}{2.13.15}=\dfrac{33}{65}\Rightarrow \widehat{B}\approx {{59}^{0}}29'$

Ta có $a=\sqrt{{{b}^{2}}+{{c}^{2}}-2bc\cos A}=\sqrt{{{6}^{2}}+{{8}^{2}}-2.6.8.\cos 60{}^\circ }=2\sqrt{13}$

 Ta có : $S=\dfrac{1}{2}b.c.\sin A\Rightarrow \sin A=\dfrac{2S}{b.c}=\dfrac{2.3\sqrt{3}}{4.3}=\dfrac{\sqrt{3}}{2}$ $\Rightarrow \left[ \begin{array}{l}   & A=60{}^\circ \\   & A=120{}^\circ \\  \end{array} \right.$ .
 Do góc $A$ tù nên $A=120{}^\circ$ .
Khi đó: ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc.\cos A={{4}^{2}}+{{3}^{2}}-2.4.3.\left( -\dfrac{1}{2} \right)=37$
$\Rightarrow a=\sqrt{37}$ .

Ta có: ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2a.b.\cos C={{8}^{2}}+{{10}^{2}}-2.8.10.\cos {{60}^{0}}=84\Rightarrow c=2\sqrt{21}$.

$\begin{array}{l} +) A + B = {180^0} - C\\  \Rightarrow A + B - 2C = {180^0} - 3C\\  \Rightarrow \sin \left( {A + B - 2C} \right) = \sin \left( {{{180}^0} - 3C} \right) = \sin 3C\\  +) \cos \frac{{B + C}}{2} = \cos \left( {{{90}^0} - \frac{A}{2}} \right) = \sin \frac{A}{2}\\  +) \sin \left( {A + B} \right) = \sin \left( {{{180}^0} - C} \right) = \sin C \end{array}$

Có $\dfrac{A+B+2C}{2}={{90}^{0}}+\dfrac{C}{2}$

$\Rightarrow \cos \left( \dfrac{A+B+2C}{2} \right)=\cos \left( {{90}^{0}}+\dfrac{C}{2} \right)$

$\Leftrightarrow \cos \left( \dfrac{A+B+2C}{2} \right)=-\sin \dfrac{C}{2}$

Ta có ${{a}^{2}}+{{b}^{2}}={{6}^{2}}+{{8}^{2}}=100={{c}^{2}}$ nên tam giác $ABC$ vuông tại $C$ .
Do đó diện tích tam giác $ABC$ là $S=\dfrac{1}{2}.a.b=\dfrac{1}{2}.6.8=24$ .