1) Diện tích tam giác \(S = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}.35.20.\sin {60^0} = 140\sqrt 3 \)
Áp dụng định lí cos ta có:
\(\begin{array}{l}{a^2} = {b^2} + {c^2} - 2bc\cos A\\ = {20^2} + {35^2} - 2.20.35\cos {60^0}\\ = 925\\ \Rightarrow a = 5\sqrt {37} \end{array}\)
Ta có: \(S = \dfrac{1}{2}a{h_a} \Leftrightarrow 140\sqrt 3 = \dfrac{1}{2}.5\sqrt {37} .{h_a}\) \( \Leftrightarrow {h_a} = \dfrac{{2.140\sqrt 3 }}{{5\sqrt {37} }} = \dfrac{{56\sqrt 3 }}{{\sqrt {37} }}\)
2) Ta có: \(\dfrac{a}{{\sin A}} = 2R \Leftrightarrow \dfrac{{5\sqrt {37} }}{{\sin 60}} = 2R \Leftrightarrow R = \dfrac{{5\sqrt {37} }}{{2\sin 60}} = \dfrac{{5\sqrt {37} }}{{\sqrt 3 }}\)
3) \(S = pr \Leftrightarrow r = \dfrac{S}{p} = \dfrac{{140\sqrt 3 }}{{\dfrac{{20 + 35 + 5\sqrt {37} }}{2}}} = \dfrac{{280\sqrt 3 }}{{55 + 5\sqrt {37} }} = \dfrac{{56\sqrt 3 }}{{11 + \sqrt {37} }}\)
