§3. Các hệ thức lượng giác trong tam giác và giải tam giác

Ta có:

\(\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}\)

\(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{DC}\)

Để \(BD\perp AM\Rightarrow\overrightarrow{DB}.\overrightarrow{AM}=0\)

\(\Rightarrow\left(\overrightarrow{DA}+\overrightarrow{AB}\right)\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{DC}\right)=0\)

\(\Leftrightarrow\overrightarrow{DA}.\overrightarrow{AB}+\overrightarrow{DA}.\overrightarrow{AD}+\overrightarrow{DA}.\overrightarrow{DC}+\overrightarrow{AB}^2+\overrightarrow{AB}.\overrightarrow{AD}+\overrightarrow{AB}.\overrightarrow{DC}=0\)

\(\Leftrightarrow0-h^2+0+a^2+0+ab=0\)

\(\Leftrightarrow a^2-h^2+ab=0\)

Đây là hệ thức cần tìm

giải nhanh giúp mik vs

\(A=sinx\cdot cosx\cdot\dfrac{sinx}{cosx}+sinx\cdot cosx\cdot\dfrac{cosx}{sinx}\)

\(=sin^2x+cos^2x=1\)

\(B=\dfrac{sinx+sinx\cdot cosx+sinx-sinx\cdot cosx}{sin^2x}=\dfrac{2}{sinx}\)

\(=\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a\left(sin^2a+cos^2a\right)+sin^2a\cdot cos^2a\)

\(=1-sin^2a\cdot cos^2a=1-\dfrac{1}{4}sin^22a\)

Bài 1: 

\(cosa=\sqrt{1-\dfrac{5}{9}}=\dfrac{2}{3}\)

\(tana=\dfrac{\sqrt{5}}{3}:\dfrac{2}{3}=\dfrac{\sqrt{5}}{2}\)

\(cota=1:\dfrac{\sqrt{5}}{2}=\dfrac{2}{\sqrt{5}}\)

\(\widehat{A}=180^0-\left(\widehat{B}+\widehat{C}\right)=105^0\)

Áp dụng định lý hàm sin:

\(\dfrac{BC}{sin\widehat{A}}=\dfrac{AC}{sin\widehat{B}}=\dfrac{AB}{sin\widehat{C}}\Rightarrow\left\{{}\begin{matrix}AB=\dfrac{BC.sin\widehat{C}}{sin\widehat{A}}\approx14,64\left(cm\right)\\AC=\dfrac{BC.sin\widehat{B}}{sin\widehat{A}}\approx10,35\left(cm\right)\end{matrix}\right.\)

Xét `\triangle ABC` có: `\hat{A}=180^o -\hat{B}-\hat{C}`

                                                `=180^o -30^o -60^o =90^o`

   `=>\triangle ABC` vuông tại `A`

`@a/[sin A]=2R<=>5/[sin 90^o]=2R<=>R=2,5`

`@sin B=b/a<=>sin 30^o=b/5=>b=2,5`

`S=1/2ab sin C=1/2 . 5.2,5.sin 60^o=[25\sqrt{3}]/8`

Xét ΔABC có \(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

=>\(\dfrac{6^2+BC^2-8^2}{2\cdot6\cdot BC}=-\dfrac{1}{2}\)

=>\(\dfrac{BC^2-28}{12BC}=\dfrac{-1}{2}\)

=>\(2BC^2-56=-12BC\)

=>2BC^2+12BC-56=0

=>\(BC=-3+\sqrt{37}\left(cm\right)\)