Question

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Answer 1

$\it \Delta ABC\ dreprunghic\ \hat\imath n\ B\ \Rightarrow tg(\widehat{BCA})=\dfrac{4\sqrt6}{4\sqrt2}=\sqrt3\ \ \ \ (1)\\ \\ \\ Dar,\ \ tg60^0=\sqrt3\ \ \ \ \ (2)\\ \\ \\ (1),\ (2)\ \Rightarrow \widehat{BCA}=60^0\ \ \ \ (3)\\ \\ \\ \Delta BOC-isoscel,\ OB=OC\ \ \ \ (4)$

$\it (3),\ (4)\ \Rightarrow \Delta BOC-\ echilateral\ \Rightarrow \widehat{BOC}=60^0\\ \\ \\ \widehat{AOD}=\widehat{BOC}=60^0\ \ (opuse\ la\ v\hat{a}rf)\\ \\ \\ sin(AOD)=sin60^0=\dfrac{\sqrt3}{2}$