Bài tập 6.31 trang 190 SBT Toán 10

Ta có:

\(\begin{array}{l}
\cos \alpha  = \sqrt {1 - \frac{{64}}{{289}}}  = \frac{{15}}{{17}}\\
\cos \beta  = \sqrt {1 - \frac{{225}}{{289}}}  = \frac{8}{{17}}
\end{array}\)

Do đó :

\(\begin{array}{l}
\sin \left( {\alpha  + \beta } \right) = \sin \alpha \cos \beta  + \cos \alpha \sin \beta \\
 = \frac{8}{{17}}.\frac{8}{{17}} + \frac{{15}}{{17}}.\frac{{15}}{{17}} = \frac{{289}}{{289}} = 1
\end{array}\)

Vì \(0 < \alpha  < \frac{\pi }{2},0 < \beta  < \frac{\pi }{2}\) nên suy ra \(\alpha  + \beta  = \frac{\pi }{2}\)

-- Mod Toán 10 HỌC247