Đáp án:
\( - 13\cos a + 5\sin b = 16\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {\dfrac{5}{{13}}} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow \cos a = \pm \dfrac{{12}}{{13}}\\
\dfrac{\pi }{2} < a < \pi \Rightarrow \cos a < 0 \Rightarrow \cos a = - \dfrac{{12}}{{13}}\\
{\sin ^2}b + {\cos ^2}b = 1\\
\Leftrightarrow {\sin ^2}b + {\left( {\dfrac{3}{5}} \right)^2} = 1\\
\Leftrightarrow \sin b = \pm \dfrac{4}{5}\\
0 < b < \dfrac{\pi }{2} \Rightarrow \sin b > 0 \Rightarrow \sin b = \dfrac{4}{5}\\
\Rightarrow - 13\cos a + 5\sin b = - 13.\left( { - \dfrac{{12}}{{13}}} \right) + 5.\dfrac{4}{5} = 12 + 4 = 16
\end{array}\)
