Đề sai hết ở cả hai câu rồi bạn

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:

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

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

+) có AB²+AC²=9²+12²=225
        và BC²=15²=225
=> AB²+AC²=BC²
=> tam giac ABC vuong tai A
+) góc A=90^o
sin B= AC/BC=12/15 => B=53^o
=> goc C=180-90-53=37^o

Bài 2: 

a: \(BC=\sqrt{8^2+6^2}=10\left(cm\right)\)

b: \(BC=\sqrt{12^2+16^2}=20\left(cm\right)\)

c: \(BC=\sqrt{5^2+12^2}=13\left(cm\right)\)

a) Áp dụng định lý Py-tago ta có:

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

\(\Rightarrow AB=\sqrt{BC^2-AC^2}=\sqrt{15^2-10^2}=5\sqrt{5}\left(cm\right)\)

b) Áp dụng định lý Py-tago ta có:

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

\(\Rightarrow BC=\sqrt{AC^2+AB^2}=\sqrt{12^2+7^2}=\sqrt{193}\left(cm\right)\)

c) Ta có: \(cosB=\dfrac{AB}{BC}\)

\(\Rightarrow BC=\dfrac{AB}{cosB}=\dfrac{7}{cos50^o}\approx11\left(cm\right)\)

Áp dụng định lý Py-tago ta có:

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

\(\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{11^2-7^2}=6\sqrt{2}\left(cm\right)\)

d) Ta có:

\(\widehat{B}+\widehat{C}=90^o\Rightarrow\widehat{B}=90^o-65=25^o\)

Mà: \(sinB=\dfrac{AC}{BC}\)

\(\Rightarrow AC=sinB\cdot BC=sin25^o\cdot10\approx4\left(cm\right)\)

Áp dụng định lý Py-tago ta có:

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

\(\Rightarrow AB=\sqrt{BC^2-AC^2}=\sqrt{10^2-4^2}=2\sqrt{21}\left(cm\right)\)