bạn tự vẽ hình nha thông cảm cho mình
a) vẽ đường cao BH (BH⊥AC,H∈AC)
Ta có : \(\sin A+\cos A=\frac{BH}{AB}+\frac{AH}{AB}\)\(\left(\sin A=\frac{BH}{AB},\cos A=\frac{AH}{AB}\right)\)
\(\Leftrightarrow\sin A+\cos A=\frac{BH+AH}{AB}\)
Xét tam giác AHB ta có : \(BH+AH>AB\) (BĐT tam giác)
\(\Leftrightarrow\)\(\frac{BH+AH}{AB}>1\)
\(\Leftrightarrow\sin A+cosA>1\)(đpcm)
b)Ta có :\(\cot B=\frac{BH}{AH},\cot C=\frac{HC}{AH},BH+HC=BC\)
VP:\(AH\cdot\left(\cot B+\cot C\right)\)
\(=AH\cdot\left(\frac{BH}{AH}+\frac{HC}{AH}\right)\)
\(=BH+HC\)
\(=BC\) (đpcm)
c) Ta có:\(\tan B=\frac{AH}{BH}\)
Hay \(\tan\left(60\right)=\frac{6}{BH}\)
\(\Leftrightarrow BH=\frac{6}{\tan\left(60\right)}\)
\(\Leftrightarrow BH=2\sqrt{3}\)
Ta có :\(\tan\left(45\right)=\frac{AH}{HC}\)
Hay \(\tan\left(45\right)=\frac{6}{HC}\)
\(\Leftrightarrow HC=\frac{6}{\tan\left(45\right)}\)
\(\Leftrightarrow HC=6\)
Ta có :BH+HC=BC
Hay \(2\sqrt{3}+6=BC\)
\(\Leftrightarrow2\sqrt{3}+6\approx9.5\)
Ta có: SABC \(=\frac{1}{2}\cdot BC\cdot AH\)
Hay SABC\(=\frac{1}{2}6\cdot9.5\)
\(\Leftrightarrow SABC=28.5\)
Vậy SABC=28.5cm
