b) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow AB^2=5.7^2-4.1^2=15,68\)

hay \(AB\simeq3,96\left(cm\right)\)

Xét ΔABC vuông tại A có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{41}{57}\)

nên \(\widehat{B}\simeq46^0\)

Ta có: ΔABC vuông tại A(gt)

nên \(\widehat{B}+\widehat{C}=90^0\)(hai góc nhọn phụ nhau)

\(\Leftrightarrow\widehat{C}+46^0=90^0\)

hay \(\widehat{C}=44^0\)

a) Ta có: ΔABC vuông tại A(gt)

nên \(\widehat{C}+\widehat{B}=90^0\)(hai góc nhọn phụ nhau)

\(\Leftrightarrow\widehat{B}+60^0=90^0\)

hay \(\widehat{B}=30^0\)

Xét ΔABC vuông tại A có 

\(AC=AB\cdot\tan\widehat{B}\)

\(\Leftrightarrow AC=10\cdot\tan30^0\)

hay \(AC=\dfrac{10\sqrt{3}}{3}\left(cm\right)\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=\left(\dfrac{10\sqrt{3}}{3}\right)^2+10^2=\dfrac{400}{3}\)

hay \(BC=\dfrac{20\sqrt{3}}{3}\left(cm\right)\)

\(AB=\cos B\cdot BC=\dfrac{1}{2}\cdot20=10\left(cm\right)\\ AC=\sin B\cdot BC=\dfrac{\sqrt{3}}{2}\cdot20=10\sqrt{3}\approx17,3205\left(cm\right)\\ \widehat{C}=90^0-\widehat{B}=30^0\)

a) Ta có:

\(\widehat{B}=180^o-90^o-52^o=28^o\) 

\(sinB=\dfrac{AC}{BC}\Rightarrow sin28^o=\dfrac{AC}{12}\)

\(\Rightarrow AC=sin28^o\cdot12\approx3,25\left(cm\right)\)

Áp dụng Py-ta-go ta có:

\(AB^2=BC^2-AC^2\)

\(\Rightarrow AB=\sqrt{BC^2-AC^2}=\sqrt{12^2-3,25^2}\)

\(\Rightarrow AB\approx11,55\left(cm\right)\)

b) Áp dụng Py-ta-go ta có:

\(BC^2=AB^2+AC^2\)

\(\Rightarrow BC=\sqrt{5^2+8^2}\approx9,43\left(cm\right)\) 

Mà: \(sinB=\dfrac{AC}{BC}=\dfrac{8}{9,43}\)

\(\Rightarrow\widehat{B}\approx58^o\)

\(\Rightarrow\widehat{C}=180^o-90^o-58^o=22^o\)

c) Ta có:

\(\widehat{C}=180^o-90^o-35^o=55^o\)

\(sinB=\dfrac{AC}{BC}\Rightarrow sin35^o=\dfrac{10}{BC}\)

\(\Rightarrow BC=\dfrac{10}{sin35^o}\approx17,43\left(cm\right)\)

Áp dụng Py-ta-go ta có:

\(AB^2=BC^2-AC^2\)

\(\Rightarrow AB=\sqrt{17,43^2-10^2}\approx14,27\left(cm\right)\)

a) Áp dụng HTL :

\(\left\{{}\begin{matrix}AH^2=BH.HC\Rightarrow AH=\sqrt{1,8.3,2}=2,4\left(cm\right)\\AB^2=BH.BC\Rightarrow AB=\sqrt{1,8\left(1,8+3,2\right)}=3\left(cm\right)\\AC^2=HC.BC\Rightarrow AC=\sqrt{3,2\left(1,8+3,2\right)}=4\left(cm\right)\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}tanB=\dfrac{AC}{AB}=\dfrac{4}{3}\Rightarrow\widehat{B}\approx53^0\\tanC=\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow\widehat{C}\approx37^0\end{matrix}\right.\)