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a. Ta thấy \(\left(a\sqrt{5}\right)^2=\left(a\sqrt{3}\right)^2+\left(a\sqrt{2}\right)^2\Rightarrow AB^2=BC^2+AC^2\)
\(\Rightarrow\Delta ABC\)vuông tại C
b. \(\sin B=\frac{AC}{AB}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5};\cos B=\frac{CB}{AB}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{15}}{5}\)
\(\tan B=\frac{AC}{AB}=\frac{\sqrt{6}}{3};\cot B=\frac{\sqrt{6}}{2}\)
\(\sin A=\cos B=\frac{\sqrt{15}}{5};\cos A=\sin B=\frac{\sqrt{10}}{5}\)
\(\tan A=\cot B=\frac{\sqrt{6}}{2};\cot A=\tan B=\frac{\sqrt{6}}{3}\)
\(\cos C=\sqrt{1-\sin^2C}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}\)
\(\Rightarrow\cos C=\frac{4}{5}\)
\(\Rightarrow\tan C=\frac{\sin C}{\cos C}=\frac{3}{5}:\frac{4}{5}=\frac{3}{4}\)và \(\cot C=\frac{4}{3}\)
Ta có: \(\widehat{C};\widehat{B}\)là hai góc phụ nhau
\(\Rightarrow\hept{\begin{cases}\sin C=\cos B\\\cos C=\sin B\end{cases};\hept{\begin{cases}\tan C=\cot B\\\cot C=\tan B\end{cases}}}\)
\(\Rightarrow\sin B=\frac{4}{5};\cos B=\frac{3}{5};\tan B=\frac{4}{3};\cot B=\frac{3}{4}\)
Ta có: \(\sin C=\frac{AB}{BC}=\frac{3}{5}\)
=> \(\frac{AB}{3}=\frac{BC}{5}=k\left(k\inℕ\right)\)
=> \(\hept{\begin{cases}AB=3k\\BC=5k\end{cases}}\)
=> \(AC=\sqrt{\left(5k\right)^2-\left(3k\right)^2}=\sqrt{16k^2}=4k\)
Đến đây thì xong rồi:))
\(\sin B=\frac{AC}{BC}=\frac{4k}{5k}=\frac{4}{5}\) ; \(\cos B=\frac{AB}{BC}=\frac{3k}{5k}=\frac{3}{5}\)
\(\tan B=\frac{AC}{AB}=\frac{4k}{3k}=\frac{4}{3}\) ; \(\cot B=\frac{AB}{AC}=\frac{3k}{4k}=\frac{3}{4}\)
B A C H
Xét \(\Delta ABC\)có \(AH^2=BH.CH=25.64=1600\Rightarrow AH=40\left(cm\right)\)
\(AC^2=CH.BC=64.\left(64+25\right)=5696\Rightarrow AC=8\sqrt{89}\left(cm\right)\)
\(AB^2=BH.BC=25.89=2225\Rightarrow AB=5\sqrt{89}\left(cm\right)\)
Ta có \(\sin B=\frac{AC}{BC}=\frac{8\sqrt{89}}{89}\Rightarrow\widehat{B}\approx58^0\)\(\Rightarrow\widehat{C}=180^0-\widehat{A}-\widehat{B}=180^0-90^0-58^0=32^0\)
Bài 2:
Xét \(\Delta ABC\)có \(\widehat{A}=90^o\)và\(AH\perp BC\)
\(\Rightarrow AH^2=HB.HC\)(Hệ thức lượng)
\(AH^2=25.64\)
\(AH=\sqrt{1600}=40cm\)
Xét \(\Delta ABH\)có\(\widehat{H}=90^o\)
\(\Rightarrow\tan B=\frac{AH}{BH}\)\(=\frac{40}{25}=\frac{8}{5}\)
\(\Rightarrow\widehat{B}\approx58^o\)
Xét \(\Delta ABC\)có \(\widehat{A}=90^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=90^o\)
\(58^o+\widehat{C}=90^o\)
\(\Rightarrow\widehat{C}\approx90^o-58^o\)
\(\widehat{C}\approx32^o\)
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