Đáp án: $\widehat{BIC}=57.5^o$
Giải thích các bước giải:
Ta có:
$\widehat{BIC}=180^o-\widehat{IBC}-\widehat{ICB}$
$\to \widehat{BIC}=180^o-\dfrac12\widehat{xBC}-\dfrac12\widehat{yCB}$
$\to \widehat{BIC}=90^o-\dfrac12\widehat{xBC}+90^o-\dfrac12\widehat{yCB}$
$\to \widehat{BIC}=\dfrac12(180^o-\widehat{xBC})+\dfrac12(180^o-\widehat{yCB})$
$\to \widehat{BIC}=\dfrac12\widehat{ABC}+\dfrac12\widehat{ACB}$
$\to \widehat{BIC}=\dfrac12(\widehat{ABC}+\widehat{ACB})$
$\to \widehat{BIC}=\dfrac12(180^o-\widehat{BAC})$
$\to \widehat{BIC}=57.5^o$