Question

Cine poate sa ma ajute la acest Exercițiu? Va rog frumos!!!!

Answer 1

$\displaystyle\\a)~\mathcal{A}_{\textnormal{ABC}}=\frac{\textnormal{AC}\cdot \textnormal{d(B,AC)}}{2} \Longleftrightarrow \textnormal{d(B,AC)}=\frac{2\mathcal{A}_{\textnormal{ABC}}}{\textnormal{AC}}=2\sqrt{2}\textnormal{cm}.\\\\b)~\sin(\measuredangle\textnormal{BAC})=\frac{\textnormal{d(B,AC)}}{\textnormal{AB}}=\frac{\sqrt{2}}{2},~\textnormal{de~unde,}~\measuredangle\textnormal{BAC}=45^{\circ} \textnormal{~sau}~\measuredangle\textnormal{BAC} =135^{\circ}.$

$\displaystyle\\c)~\textnormal{Triunghiul~ABC~este~isoscel}\implies \measuredangle\textnormal{ACB}=\measuredangle\textnormal{ABC}.$

$\displaystyle\\~\measuredangle\textnormal{BAC}+2\measuredangle\textnormal{ACB}=180^{\circ} \Longrightarrow \measuredangle\textnormal{ACB}=\frac{180^{\circ}-\measuredangle\textnormal{BAC}}{2}=90^{\circ}-\frac{1}{2}\measuredangle\textnormal{BAC}.\\\textnormal{Daca~:}~\measuredangle\textnormal{BAC}=135^{\circ}\Longrightarrow \measuredangle\textnormal{ACB}=22^{\circ}30'.$

$\displaystyle\\\textnormal{Daca~:}~\measuredangle\textnormal{BAC}=45^{\circ} \implies \measuredangle\textnormal{ACB}=67^{\circ}30'.$

Answer 2

Răspuns:

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