1.
\(sinB=\dfrac{AC}{BC}\Rightarrow AC=BC.sinB\)
\(\Rightarrow AC^2=BC^2.\left(sinB\right)^2\)
\(\Rightarrow AC^2=\left(AB^2+AC^2\right).\left(sinB\right)^2\)
\(\Rightarrow AC^2=\left(4+AC^2\right).\dfrac{1}{9}\)
\(\Rightarrow8AC^2=4\)
\(\Rightarrow AC=\dfrac{\sqrt{2}}{2}\)
Cách 2: \(sin^2B+cos^2B=1\Leftrightarrow\dfrac{1}{9}+cos^2B=1\)
\(\Rightarrow cos^2B=\dfrac{8}{9}\Rightarrow cosB=\dfrac{2\sqrt{2}}{3}\)
\(\Rightarrow tanB=\dfrac{sinB}{cosB}=\dfrac{\sqrt{2}}{4}\)
Ta có: \(tanB=\dfrac{AC}{AB}\Rightarrow AC=AB.tanB=2.\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{2}\)
2.
\(sin^2x+cos^2x=1\)
\(\Leftrightarrow sin^2x+\left(\dfrac{1}{4}\right)^2=1\)
\(\Leftrightarrow sin^2x=\dfrac{15}{16}\)
\(\Rightarrow sinx=\dfrac{\sqrt{15}}{4}\)
\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\sqrt{15}\)
3.
\(tanC=\dfrac{AB}{AC}\Rightarrow AB=AC.tanC\)
\(\Rightarrow AB^2=AC^2.\left(tanC\right)^2=4AC^2\)
\(\Rightarrow BC^2-AC^2=4AC^2\)
\(\Rightarrow AC^2=\dfrac{BC^2}{5}=\dfrac{81}{5}\)
\(\Rightarrow AC=\dfrac{9\sqrt{5}}{5}\)
\(AB=AC.tanC=\dfrac{18\sqrt{5}}{5}\)
4.
\(tanB=\dfrac{AH}{BH}\Rightarrow BH=\dfrac{AH}{tanB}=6\)
Do \(B+C=90^0\Rightarrow tanB=cotC\)
\(cotC=\dfrac{CH}{AH}\Rightarrow CH=AH.cotC=\dfrac{2}{3}\)
\(\Rightarrow BC=BH+CH=\dfrac{20}{3}\)
Áp dụng hệ thức lượng:
\(AB^2=BH.BC\Rightarrow AB=\sqrt{BH.BC}=2\sqrt{10}\)
\(AC=\sqrt{CH.BC}=\dfrac{2\sqrt{10}}{3}\)
5.
\(1+tan^2x=1+\dfrac{sin^2x}{cos^2x}=\dfrac{cos^2x+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}\) (đpcm)
6.
Trong tam giác vuông ACH:
\(tanC=\dfrac{AH}{CH}\Rightarrow AH=CH.tanC=\dfrac{2}{3}\)
Áp dụng định lý Pitago:
\(AC=\sqrt{AH^2+CH^2}=\dfrac{2\sqrt{10}}{3}\)
Trong tam giác vuông ABH:
\(AB=\sqrt{BH^2+AH^2}=\dfrac{2\sqrt{37}}{3}\)
7.
Trong tam giác vuông ABH:
\(cosB=\dfrac{BH}{AB}\Rightarrow AB=\dfrac{BH}{cosB}=\dfrac{15}{2}\)
Áp dụng định lý Pitago:
\(AH=\sqrt{AB^2-BH^2}=\dfrac{3\sqrt{21}}{2}\)
Áp dụng định lý Pitago cho tam giác vuông ACH:
\(AC=\sqrt{AH^2+CH^2}=\dfrac{17}{2}\)
8.
Áp dụng định lý Pitago cho tam giác vuông ABH:
\(AB=\sqrt{AH^2+BH^2}=2\sqrt{5}\)
Trong tam giác vuông ACH:
\(cosC=\dfrac{CH}{AC}\Rightarrow CH=AC.cosC=\dfrac{AC}{3}\)
\(\Leftrightarrow CH^2=\dfrac{AC^2}{9}\)
\(\Leftrightarrow AC^2-AH^2=\dfrac{AC^2}{9}\)
\(\Rightarrow AC^2=\dfrac{9}{8}AH^2=18\)
\(\Rightarrow AC=3\sqrt{2}\)
9.
\(tanx=3\Leftrightarrow tan^2x=9\)
\(\Leftrightarrow\dfrac{sin^2x}{cos^2x}=9\)
\(\Leftrightarrow\dfrac{sin^2x}{1-sin^2x}=9\)
\(\Leftrightarrow sin^2x=\dfrac{9}{10}\)
\(\Leftrightarrow sinx=\dfrac{3\sqrt{10}}{10}\)
\(cosx=\dfrac{sinx}{tanx}=\dfrac{\sqrt{10}}{10}\)
Câu 10 đề sai, trong tam giác vuông BCK thì BC là cạnh huyền, BK là cạnh góc vuông thì BC không thể nhỏ hơn BK
