Bài 2: 

b: \(AH\cdot\left(\cot\widehat{B}+\cot\widehat{C}\right)\)

\(=AH\cdot\left(\dfrac{BH}{AH}+\dfrac{CH}{AH}\right)\)

\(=AH\cdot\dfrac{BC}{AH}=BC\)

a.

Trong tam giác vuông ABH ta có:

\(cotB=\dfrac{BH}{AH}\Rightarrow BH=AH.cotB\)

Trong tam giác vuông ACH ta có:

\(cotC=\dfrac{CH}{AH}\Rightarrow CH=AH.cotC\)

\(\Rightarrow BH+CH=AH.cotB+AH.cotC\)

\(\Leftrightarrow BC=AH\left(cotB+cotC\right)\)

\(\Leftrightarrow AH=\dfrac{BC}{cotB+cotC}\) (đpcm)

b. Áp dụng công thức câu a:

\(AH=\dfrac{4}{cot45^0+cot30^0}=-2+2\sqrt{3}\) (cm)

\(S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.\left(-2+2\sqrt{3}\right).4=-4+4\sqrt{3}\approx2,93\left(cm^2\right)\)

a) Kẻ đường cao BK

 Ta có: 

\(\sin\widehat{A}=\frac{BK}{AB};\cos\widehat{A}=\frac{AK}{AB}\)

=> \(\sin\widehat{A}+\cos\widehat{A}=\frac{BK}{AB}+\frac{AK}{AB}=\frac{AK+BK}{AB}>\frac{AB}{AB}=1\)

b) Kẻ đường cao AH.

Đặt BH = x => HC = a - x.  

+) Tam giác AHB vuông cân => AH = BH =x  (1) 

+) Tam giác AHC có \(\tan\widehat{ACH}=\frac{AH}{HC}\Rightarrow\tan60^o=\frac{AH}{a-x}\Rightarrow AH=\sqrt{3}\left(a-x\right)\) (2)

Từ (1) ; (2) => \(x=\sqrt{3}\left(a-x\right)\Rightarrow x=\frac{\sqrt{3}a}{1+\sqrt{3}}\)

=> \(AH=\frac{\sqrt{3}a}{1+\sqrt{3}}\)

=> \(S_{\Delta ABC}=\frac{1}{2}AH.BC=\frac{1}{2}.\frac{\sqrt{3}a}{1+\sqrt{3}}.a=\frac{3-\sqrt{3}}{4}a^2\)

Kẻ đường cao AH ứng với BC

Trong tam giác vuông ABH ta có:

\(cotB=\dfrac{BH}{AH}\Rightarrow BH=AH.cotB\)

Trong tam giác vuông ACH ta có:

\(cotC=\dfrac{CH}{AH}\Rightarrow CH=AH.cotC\)

\(\Rightarrow BH+CH=AH.cotB+AH.cotC\)

\(\Leftrightarrow BC=AH\left(cotB+cotC\right)\)

\(\Leftrightarrow AH=\dfrac{BC}{cotB+cotC}\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.\dfrac{BC^2}{cotB+cotC}=\dfrac{\left(2a\right)^2}{2\left(cot45^0+cot60^0\right)}=\left(3-\sqrt{3}\right)a^2\)