a: Diện tích tam giác ABC là:\(S_{ABC}=\frac12\cdot BA\cdot BC\cdot\sin ABC\) \(=\frac12\cdot10\cdot14\cdot\sin60=5\cdot14\cdot\frac{\sqrt3}{2}=\frac{70\sqrt3}{2}=35\sqrt3\left(\operatorname{cm}^2\right)\) Xét ΔBAC có \(cosB=\frac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\) =>\(\frac{10^2+14^2-AC^2}{2\cdot10\cdot14}=cos60=\frac12\) =>\(100+196-AC^2=10\cdot14=140\) =>\(AC^2=100+196-140=196-40=156\) =>\(AC=2\sqrt{39}\left(\operatorname{cm}\right)\) Chu vi tam giác ABC là:AB+BC+AC=10+14+\(2\sqrt{39}=24+2\sqrt{39}\) (cm)b: Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\) =>\(\hat{A}=180^0-20^0-16^0=144^0\) Xét ΔABC có \(\frac{BC}{\sin A}=\frac{AC}{sinB}=\frac{AB}{\sin C}\) =>\(\frac{AC}{\sin20}=\frac{AB}{\sin16}=\frac{12}{\sin144}\) =>AC≃7(cm); AB≃5,6(cm)Chu vi tam giác ABC là:AB+AC+BC=7+5,6+12=19+5,6=24,6(cm)Diện tích tam giác ABC là:\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\) \(=\frac12\cdot7\cdot5,6\cdot\sin144\) ≃11,5\(\left(\operatorname{cm}^2\right)\)Câu trả lời: a: Diện tích tam giác ABC là:\(S_{ABC}=\frac12\cdot BA\cdot BC\cdot\sin ABC\) \(=\frac12\cdot10\cdot14\cdot\sin60=5\cdot14\cdot\frac{\sqrt3}{2}=\frac{70\sqrt3}{2}=35\sqrt3\left(\operatorname{cm}^2\right)\) Xét ΔBAC có \(cosB=\frac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\) =>\(\frac{10^2+14^2-AC^2}{2\cdot10\cdot14}=cos60=\frac12\) =>\(100+196-AC^2=10\cdot14=140\) =>\(AC^2=100+196-140=196-40=156\) =>\(AC=2\sqrt{39}\left(\operatorname{cm}\right)\) Chu vi tam giác ABC là:AB+BC+AC=10+14+\(2\sqrt{39}=24+2\sqrt{39}\) (cm)b: Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\) =>\(\hat{A}=180^0-20^0-16^0=144^0\) Xét ΔABC có \(\frac{BC}{\sin A}=\frac{AC}{sinB}=\frac{AB}{\sin C}\) =>\(\frac{AC}{\sin20}=\frac{AB}{\sin16}=\frac{12}{\sin144}\) =>AC≃7(cm); AB≃5,6(cm)Chu vi tam giác ABC là:AB+AC+BC=7+5,6+12=19+5,6=24,6(cm)Diện tích tam giác ABC là:\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\) \(=\frac12\cdot7\cdot5,6\cdot\sin144\) ≃11,5\(\left(\operatorname{cm}^2\right)\)