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