Giải thích các bước giải:
Ta có:
\(\begin{array}{l}{S_{ABC}} = \frac{1}{2}.b.c.\sin A = \frac{1}{2}.8.5.\sin 60^\circ = 10\sqrt 3 \left( {c{m^2}} \right)\\{a^2} = {b^2} + {c^2} - 2.b.c.\cos A\\ \Leftrightarrow {a^2} = {8^2} + {5^2} - 2.8.5.\cos 60^\circ \\ \Leftrightarrow {a^2} = 49\\ \Leftrightarrow a = 7\left( {cm} \right)\\{S_{ABC}} = \frac{{a.b.c}}{{4R}} \Leftrightarrow R = \frac{{abc}}{{4{S_{ABC}}}} = \frac{{7.8.5}}{{4.10\sqrt 3 }} = \frac{{7\sqrt 3 }}{3}\left( {cm} \right)\\{S_{ABC}} = \frac{{a + b + c}}{2}.r \Leftrightarrow r = \frac{{2{S_{ABC}}}}{{a + b + c}} = \frac{{2.10\sqrt 3 }}{{7 + 8 + 5}} = \sqrt 3 \left( {cm} \right)\\{h_a} = \frac{{2{S_{ABC}}}}{a} = \frac{{2.10\sqrt 3 }}{7} = \frac{{20\sqrt 3 }}{7}\left( {cm} \right)\\{m_a}^2 = \frac{{{b^2} + {c^2}}}{2} - \frac{{{a^2}}}{4}\\ \Leftrightarrow {m_a}^2 = \frac{{{8^2} + {5^2}}}{2} - \frac{{{7^2}}}{4}\\ \Leftrightarrow {m_a}^2 = \frac{{129}}{4}\\ \Leftrightarrow {m_a} = \frac{{\sqrt {129} }}{2}\left( {cm} \right)\end{array}\)
Cho mình hỏi con đường tín hiệu của hệ dẫn truyền sắp xếp như nào ạ? Mình có bó His, nút xoang nhĩ, nút nhĩ thất, mạng puôckin
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