Bài 4 :
\(sinx=cosx\)
\(\Leftrightarrow cos\left(\dfrac{\pi}{2}-x\right)=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}-x=x+k2\pi\\\dfrac{\pi}{2}-x=-x+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}-k2\pi\\0x=-\dfrac{\pi}{2}+k2\pi\left(vo.ly\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}-k2\pi\left(k\in Z\right)\)
Bài 1 : Sửa lại đường cao BH thành BK
a) \(sin\widehat{B}=sin50^o=\dfrac{AH}{AB}\Rightarrow AH=AB.sin50^o=6,8.0,8=5,44\left(cm\right)\)
\(sin\widehat{A}=sin70^o=\dfrac{BK}{AB}\Rightarrow BK=AB.sin70^o=6,8.0,9=6,12\left(cm\right)\)
b) \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{C}=180^o-\left(\widehat{A}+\widehat{B}\right)=180^o-\left(70^o+50^o\right)=60^o\)
\(tan\widehat{C}=tan60^o=\dfrac{AH}{HC}\Rightarrow HC=\dfrac{AH}{tan60^o}=\dfrac{5,44}{1,73}=3,14\left(cm\right)\)
c) \(tan\widehat{B}=tan50^o=\dfrac{AH}{BH}\Rightarrow BH=\dfrac{AH}{tan50^o}=\dfrac{5,44}{1,19}=4,57\left(cm\right)\)
\(BC=BH+HC=4,57+3,14=7,71\left(cm\right)\)
\(S_{ABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.5,44.7,71\approx20,97\left(cm^2\right)\)
Bài 2: a: Xét ABC vuông tại A có \(tanC=\dfrac{AB}{AC}\)
=>\(AC=\dfrac{9}{tan30}=9\sqrt{3}\left(cm\right)\)
ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)
=>\(BC=\sqrt{9^2+\left(9\sqrt{3}\right)^2}=18\left(cm\right)\)
ΔABC vuông tại A
=>\(\widehat{ABC}+\widehat{ACB}=90^0\)
=>\(\widehat{ABC}=90^0-30^0=60^0\)
b: Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AH\cdot BC=AB\cdot AC\\BH\cdot BC=BA^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AH=\dfrac{9\cdot9\sqrt{3}}{18}=\dfrac{9\sqrt{3}}{2}\left(cm\right)\\BH=\dfrac{9^2}{18}=4,5\left(cm\right)\end{matrix}\right.\)
c: Xét ΔABC có AD là phân giác
nên \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)
=>\(AD=\dfrac{2\cdot9\cdot9\sqrt{3}}{9+9\sqrt{3}}\cdot cos45=\dfrac{18\cdot9\sqrt{3}}{9\left(1+\sqrt{3}\right)}\cdot\dfrac{\sqrt{2}}{2}=\left(27-9\sqrt{3}\right)\cdot\dfrac{\sqrt{2}}{2}\left(cm\right)\)
Bài 2 :
a) \(tanC=\dfrac{AB}{AC}\Rightarrow AC=\dfrac{AB}{tanC}=\dfrac{9}{tan30^o}=\dfrac{9}{\dfrac{\sqrt[]{3}}{3}}=9\sqrt[]{3}\left(cm\right)\)
\(BC^2=AB^2+AC^2=9^2+9^2.3=9^2.4\) (Pitago)
\(\Rightarrow BC=9.2=18\left(cm\right)\)
b) \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{9.9.\sqrt[]{3}}{18}=\dfrac{9\sqrt[]{3}}{2}\left(cm\right)\)
\(tanC=\dfrac{AH}{HC}\Rightarrow HC=\dfrac{AH}{tanC}=\dfrac{\dfrac{9\sqrt[]{3}}{2}}{\dfrac{\sqrt[]{3}}{3}}=\dfrac{27}{2}\left(cm\right)\)
\(BC=BH+HC\Rightarrow BH=BC-HC=18-\dfrac{27}{2}=\dfrac{9}{2}\left(cm\right)\)
