Đáp án: $0.76$

Giải thích các bước giải:

Kẻ $CD\perp AC, BD\perp AB, D\in (ABC)$

Vì $SA\perp (ABC)$

$\to SA\perp CD, SA\perp BD$

$\to CD\perp (SAC), BD\perp (SAB)$

$\to CD\perp (SAC), DB\perp (SAB)$

$\to CD\perp AN, BD\perp AM$

Lại có: $AM\perp SB, AN\perp SC$

$\to AN\perp (SCD), AM\perp (SBD)$

$\to AN\perp SD, AM\perp SD$

$\to SD\perp (AMN)$

Vì $\alpha=\widehat{SA, AMN}$

$\to 90^o-\alpha=\widehat{ASD}$

$\to \cos\widehat{ASD}=\sin\alpha=\dfrac{\sqrt{21}}7$

$\to \sin\widehat{ASD}=\sqrt{1-\cos^2\widehat{ASD}}=\sqrt{1-(\dfrac{\sqrt{21}}7)^2}=\dfrac2{\sqrt7}$

$\to \tan\widehat{ASD}=\dfrac{\sin\widehat{ASD}}{\cos\widehat{ASD}}$

$\to \dfrac{AD}{SA}=\dfrac2{\sqrt7}:\dfrac{\sqrt{21}}7=\dfrac2{\sqrt3}$

Ta có:

$BC^2=AB^2+AC^2-2AB\cdot AC\cdot \cos A=1^2+2^2-2\cdot 1\cdot 2\cdot\cos120^o=7$

$\to BC=\sqrt7$

Vì $\widehat{ACD}=\widehat{ABD}=90^o$

$\to ABDC$ nội tiếp đường tròn đường kính $AH$

$\to \dfrac{BC}{\sin \widehat{BAC}}=2R=AD$

$\to \dfrac{\sqrt7}{\sin120^o}=AD$

$\to AD= \dfrac{2\sqrt{21}}3$

$\to SA= \dfrac{2\sqrt{21}}3:\dfrac2{\sqrt3}=\sqrt{7}$

Thể tích khối chóp là:

$$\dfrac13SA\cdot S_{ACB}=\dfrac13\cdot\sqrt{7}\cdot \dfrac12\cdot 1\cdot 2\cdot \sin120^o=\dfrac{\sqrt{21}}{6} \approx 0.76$$