Giải thích các bước giải:
\(\begin{array}{l}
60,\\
\pi < a < \dfrac{{3\pi }}{2} \Rightarrow \left\{ \begin{array}{l}
\sin a < 0\\
\cos a < 0
\end{array} \right.\\
\tan a = 3 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = 3 \Leftrightarrow \sin a = 3\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {3\cos a} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{{10}}\\
\cos a < 0 \Rightarrow \cos a = - \dfrac{1}{{\sqrt {10} }}\\
62,\\
\dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\tan a = - 2 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = - 2 \Leftrightarrow \sin a = - 2\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( { - 2\cos a} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{1}{5}\\
\cos a < 0 \Rightarrow \cos a = - \dfrac{1}{{\sqrt 5 }}\\
63,\\
\cos \left( {a - b} \right) = \cos a.\cos b + \sin a.\sin b\\
64,\\
\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\\
\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}\\
65,\\
\cos 2a = 1 - 2{\sin ^2}a = 1 - 2.{\left( {\dfrac{3}{4}} \right)^2} = - \dfrac{1}{8}\\
66,\\
\widehat A = 180^\circ - \widehat B - \widehat C = 71^\circ 27'\\
\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R \Rightarrow b = \dfrac{{a.\sin B}}{{\sin A}} = 12,4\\
67,\\
\widehat C = 180^\circ - \widehat A - \widehat B = 77^\circ 4'\\
\dfrac{{BC}}{{\sin A}} = \dfrac{{AC}}{{\sin B}} = \dfrac{{AB}}{{\sin C}} \Rightarrow AC = \dfrac{{AB.\sin B}}{{\sin C}} = 68\\
68,\\
\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} = \dfrac{{33}}{{65}}\\
\Rightarrow \widehat B = 59^\circ 29'\\
69,\\
\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = - \dfrac{7}{{15}}\\
\Rightarrow \widehat A = 117^\circ 49'
\end{array}\)
