Lời giải:
a. Áp dụng định lý sin: \(\frac{BC}{\sin A}=\frac{AC}{\sin B}=\frac{AB}{\sin C}\)
\(\Leftrightarrow \frac{BC}{\sin A}=\frac{10}{\sin B}=\frac{5}{\sin 30^0}=10\)
\(\Rightarrow \sin B=1\Rightarrow \widehat{B}=90^0\)
\(\widehat{A}=180^0-(\widehat{B}+\widehat{C})=180^0-(90^0+30^0)=60^0\)
\(\frac{BC}{\sin A}=10\Rightarrow BC=10\sin A=10\sin 60^0=5\sqrt{3}\) (cm)
b.
$\widehat{A}=180^0-(\widehat{B}+\widehat{C})$
$=180^0-(60^0+45^0)=75^0$
Áp dụng định lý sin:
\(\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{CA}{\sin B}\Leftrightarrow \frac{10}{\sin 45^0}=\frac{BC}{\sin 75^0}=\frac{CA}{\sin 60^0}=10\sqrt{2}\)
\(\Rightarrow BC=10\sqrt{2}.\sin 75^0=5+5\sqrt{3}\) (cm)
\(AC=10\sqrt{2}\sin 60^0=5\sqrt{6}\) (cm)
