Theo đề bài ta có :
\(\dfrac{B}{C}=\dfrac{4}{3}=\dfrac{B}{4}=\dfrac{C}{3}=\dfrac{B+C}{4+3}=\dfrac{105}{7}=15\)
\(\dfrac{B}{4}=15\Rightarrow B=60^0\)
\(\dfrac{C}{3}=15\Rightarrow C=45^0\)
Vậy \(\left\{{}\begin{matrix}\widehat{B}=60^0\\\widehat{C}=45^0\end{matrix}\right.\)
Ta có :
\(\sin B=\dfrac{AH}{AB}\Rightarrow AH=\sin B.AB\Rightarrow AH=\sin60^0.10=5\sqrt{3}cm\)
Vậy \(AH=5\sqrt{3}cm\)
\(\cos B=\dfrac{HB}{AB}\Rightarrow HB=\cos B.AB\Rightarrow HB=\cos60^0.10=5cm\)
Vậy \(HB=5cm\)
\(\sin C=\dfrac{AH}{AC}\Rightarrow AC=\dfrac{AH}{\sin C}\Rightarrow AC=\dfrac{5\sqrt{3}}{\sin45^0}=5\sqrt{6}cm\)
Vậy \(AC=5\sqrt{6}cm\)
\(\cos C=\dfrac{HC}{AC}\Rightarrow HC=\cos C.AC\Rightarrow HC=\cos45^0.5\sqrt{6}=5\sqrt{3}cm\)
Vậy \(HC=5\sqrt{3}cm\)
Ta lại có :
\(BC=HB+HC=5+5\sqrt{3}\approx14cm\)
Vậy \(BC\approx14cm\)
Bạn tích diện tích luôn nhé
