Ta có:
\(\eqalign{
& \cos \alpha = \sqrt {1 - {{64} \over {289}}} = \sqrt {{{225} \over {289}}} = {{15} \over {17}}; \cr
& \cos \beta = \sqrt {1 - {{225} \over {289}}} = \sqrt {{{64} \over {289}}} = {8 \over {17}} \cr} \)
Do đó:
\(\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)
\({8 \over {17}}.{8 \over {17}} + {{15} \over {17}}.{{15} \over {17}} = {{289} \over {289}} = 1\)
Vì \(0 < \alpha < {\pi \over 3},0 < \beta < {\pi \over 2}\) nên từ đó suy ra \(\alpha + \beta = {\pi \over 2}\)
