Giải thích các bước giải:
Ta có $AB$ là đường kính của $(O)\to AM\perp BM$
$\to \widehat{MBA}=90^o-\widehat{BAM}=90^o-\alpha$
Ta có:
$\cos\widehat{MAB}=\dfrac{AM}{AB}$
$\to \cos\alpha=\dfrac{AM}{2R}$
$\to AM=2R\cos\alpha$
Ta có: $\widehat{MOA}= 2\widehat{BMA}=2(90^o-\alpha)$
$\to \widehat{MON}=2(90^o-\alpha)$
Mà $\sin\widehat{MON}=\dfrac{MN}{MO}$
$\to \sin(2(90^o-\alpha))=\dfrac{MN}{R}$
$\to \sin(180^o-2\alpha)=\dfrac{MN}{R}$
$\to \sin(2\alpha)=\dfrac{MN}{R}$
$\to MN=R\sin(2\alpha)$
$\to ON=\sqrt{MO^2+MN^2}$
$\to ON=R\sqrt{1+\sin^2(2\alpha)}$
$\to NA=ON-OA=R\sqrt{1+\sin^2(2\alpha)}-R=R(\sqrt{1+\sin^2(2\alpha)}-1)$
