Đáp án: $30^o$

Giải thích các bước giải:

Ta có:

$A_1D//B_1C$

$\to \widehat{A_1D, B_1I}=\widehat{IB_1C}$

Ta có:

$BD=CB_1=a\sqrt2$

$IB=IC=\dfrac12BD=\dfrac{a\sqrt2}2$

$B_1I=\sqrt{(\dfrac{a\sqrt2}2)^2+a^2}=\dfrac{\sqrt{3}a}{\sqrt{2}}$

$\to \cos\widehat{IB_1C}=\dfrac{B_1I^2+B_1C^2-IC^2}{2B_1I\cdot B_1C}=\dfrac{(\dfrac{\sqrt{3}a}{\sqrt{2}})^2+(\dfrac{a\sqrt2}2)^2-(\dfrac{a\sqrt2}2)^2}{2\cdot \dfrac{\sqrt{3}a}{\sqrt{2}}\cdot \dfrac{a\sqrt2}2}=\dfrac{\sqrt{3}}{2}$

$\to \widehat{IB_1C}=\arccos(\dfrac{\sqrt{3}}{2})=30^o$