Có:
$\left\{{}\begin{matrix}sin^2\alpha+cos^2\alpha=1\\sin\alpha=\dfrac{8}{17}\\0< \alpha< \dfrac{\pi}{2}\end{matrix}\right.$$\Leftrightarrow\left\{{}\begin{matrix}cos^2\alpha=1-\left(\dfrac{8}{17}\right)^2\\sin\alpha=\dfrac{8}{17}\\cos\alpha,sin\alpha>0\end{matrix}\right.$
$\Leftrightarrow\left\{{}\begin{matrix}cos\alpha=\dfrac{15}{17}\\sin\alpha=\dfrac{8}{17}\end{matrix}\right.$.
Tương tự: $\left\{{}\begin{matrix}sin\beta=\dfrac{15}{17}\\cos\beta=\dfrac{8}{17}\end{matrix}\right.$.
Có:$sin\left(\alpha+\beta\right)=sin\alpha cos\beta+cos\alpha sin\beta$$=\left(\dfrac{8}{17}\right)^2+\left(\dfrac{15}{17}\right)^2=1$ và $0< \alpha< \dfrac{\pi}{2};0< \beta< \dfrac{\pi}{2}$ nên: $\alpha+\beta=\dfrac{\pi}{2}$.
Cách lập luận khác: $sin\alpha=cos\beta$ và $0< \alpha< \dfrac{\pi}{2};0< \beta< \dfrac{\pi}{2}$ nên: $\alpha+\beta=\dfrac{\pi}{2}$.