Question

Answer 1

a: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA\)

\(=\dfrac{1}{2}\cdot15\cdot11\cdot sin30=\dfrac{1}{4}\cdot165=\dfrac{165}{4}\)

b: Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(15^2+11^2-BC^2=2\cdot15\cdot11\cdot cos30=165\sqrt{3}\)

=>\(BC=\sqrt{225+121-165\sqrt{3}}=\sqrt{346-165\sqrt{3}}\)

Chu vi tam giác ABC là:

\(C_{ABC}=AB+AC+BC=26+\sqrt{346-165\sqrt{3}}\)

\(S=p\cdot r\)

=>\(r=\dfrac{165}{4}:\dfrac{26+\sqrt{346-165\sqrt{3}}}{2}=\dfrac{165}{2}\left(26+\sqrt{346-165\sqrt{3}}\right)\)

Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)

=>\(2R=\sqrt{346-165\sqrt{3}}:\dfrac{1}{2}=2\cdot\sqrt{346-165\sqrt{3}}\)

=>\(R=\sqrt{346-165\sqrt{3}}\)