Centrul cercului înscris se află la intersecția bisectoarelor.
În ΔBIC:
∡BIC+∡IBC+∡ICB=180°
⇒ ∡IBC+∡ICB=180°-∡BIC=180°-123.5°=56,5°
[tex]\widehat{IBC} = \dfrac{\widehat{ABC}}{2}, \ \ \ \widehat{ICB} = \dfrac{\widehat{ACB}}{2}\\[/tex]
Atunci:
[tex]\dfrac{\widehat{ABC}}{2} + \dfrac{\widehat{ACB}}{2} = 56,5^{\circ} \implies \widehat{ABC} + \widehat{ACB} = 2 \times 56,5^{\circ} = 113^{\circ}\\[/tex]
[tex]\widehat{BAC} + \widehat{ABC} + \widehat{ACB} = 180^{\circ}[/tex]
[tex]\widehat{BAC} - (\widehat{ABC} + \widehat{ACB}) = 180^{\circ} - 113^{\circ} = \bf 67^{\circ}\\[/tex]
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Verified answer
Centrul cercului înscris se află la intersecția bisectoarelor.
În ΔBIC:
∡BIC+∡IBC+∡ICB=180°
⇒ ∡IBC+∡ICB=180°-∡BIC=180°-123.5°=56,5°
[tex]\widehat{IBC} = \dfrac{\widehat{ABC}}{2}, \ \ \ \widehat{ICB} = \dfrac{\widehat{ACB}}{2}\\[/tex]
Atunci:
[tex]\dfrac{\widehat{ABC}}{2} + \dfrac{\widehat{ACB}}{2} = 56,5^{\circ} \implies \widehat{ABC} + \widehat{ACB} = 2 \times 56,5^{\circ} = 113^{\circ}\\[/tex]
[tex]\widehat{BAC} + \widehat{ABC} + \widehat{ACB} = 180^{\circ}[/tex]
[tex]\widehat{BAC} - (\widehat{ABC} + \widehat{ACB}) = 180^{\circ} - 113^{\circ} = \bf 67^{\circ}\\[/tex]