Ta có $sin\alpha=cos(90°-\alpha)$
$\Rightarrow A=sin^210°+sin^220°+sin^230°+sin^280°+sin^270°+sin^260°\\=sin^210°+sin^220°+sin^230°+cos^210°+cos^220°+cos^230°$
Mà $sin^2\alpha +cos^2\alpha=1$
$\Rightarrow A=(sin^210°+cos^210°)+(sin^220°+cos^220°)+(sin^230°+cos^230°)\\=1+1+1\\=3$
Vậy $A=3$
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