Bài tập 35 trang 66 SGK Hình học 10 NC

a) Áp dụng định lí cosin ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
 = \frac{{{{18}^2} + {{20}^2} - {{14}^2}}}{{2.18.20}} \approx 0,73
\end{array}\\
\begin{array}{l}
\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
 = \frac{{{{14}^2} + {{20}^2} - {{18}^2}}}{{2.14.20}} \approx 0,49
\end{array}\\
{ \Rightarrow A \approx {{43}^0},B \approx {{61}^0},C \approx {{76}^0}}
\end{array}\)

b) Áp dụng định lí cosin ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
 = \frac{{{{\left( {7,3} \right)}^2} + {{\left( {4,8} \right)}^2} - {6^2}}}{{2.7,3.4,8}} \approx 0,58
\end{array}\\
\begin{array}{l}
\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
 = \frac{{{6^2} + {{\left( {4,8} \right)}^2} - {{\left( {7,3} \right)}^2}}}{{2.6.\left( {4,8} \right)}} \approx 0,1
\end{array}\\
{ \Rightarrow A \approx {{55}^0},B \approx {{85}^0},C \approx {{40}^0}}
\end{array}\)

c) Áp dụng định lí cosin ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
 = \frac{{{5^2} + {7^2} - {4^2}}}{{2.5.7}} \approx 0,83
\end{array}\\
\begin{array}{l}
\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\
 = \frac{{{4^2} + {7^2} - {5^2}}}{{2.4.7}} \approx 0,71
\end{array}\\
{ \Rightarrow A \approx {{34}^0},B \approx {{44}^0},C \approx {{102}^0}}
\end{array}\)

-- Mod Toán 10 HỌC247