Đáp án:
$SC=4a$
Giải thích các bước giải:
Ta có: $\begin{cases}AB\bot CB\text{ (do }\Delta ABC\bot B\text{)}\\AB\bot SC\text{ (do }SC\bot (ABC)\text{)}\end{cases}$
$\Rightarrow AB\bot(SBC)$
Trong $\Delta SBC$ dựng $CK\bot SB\Rightarrow CK\bot AB$
$\Rightarrow CK\bot (SAB)\Rightarrow CK\bot SA$
$\Delta SAC$ dựng $CH\bot SA$
$\Rightarrow SA\bot(CHK)\Rightarrow SA\bot HK$
$(SAB)\cap(SAC)=SA\Rightarrow\widehat{((SAB),(SAC))}=(KH,CH)=\alpha=\widehat{CHK}$
$\cos\alpha=\sqrt{\dfrac6{19}}\Rightarrow\sin^2\alpha=1-\cos^2\alpha=1-\dfrac6{19}=\dfrac{13}{19}$
$\Delta SAC\bot C:\dfrac1{CH^2}=\dfrac1{x^2}+\dfrac1{3a^2}$ $(SC=x)$
$\Rightarrow \dfrac{1}{CH^2}=\dfrac{3a^2+x^2}{x^2.3a^2}\Rightarrow CH^2=\dfrac{x^2.3a^2}{3a^2+x^2}$
$\Delta SBC\bot C:\dfrac1{CK^2}=\dfrac1{x^2}+\dfrac1{2a^2}$
$\Rightarrow\dfrac1{CK^2}=\dfrac{2a^2+x^2}{x^2.2a^2}\Rightarrow CK^2=\dfrac{x^2.2a^2}{2a^2+x^2}$
$CK\bot(SAB)\Rightarrow CK\bot HK\Rightarrow\Delta CHK\bot K$
$\sin\widehat{CHK}=\dfrac{CK}{CH}\Rightarrow\sin\alpha=\dfrac{CK}{CH}$
$\Rightarrow\sin^2\alpha=\dfrac{CK^2}{CH^2}=\dfrac{\dfrac{x^2.2a^2}{2a^2+x^2}}{\dfrac{x^2.3a^2}{3a^2+x^2}}$
$=\dfrac{x^2.2a^2}{2a^2+x^2}\dfrac{3a^2+x^2}{x^2.3a^2}$
$=\dfrac{2}{3}.\dfrac{x^2+3a^2}{x^2+2a^2}=\dfrac{13}{19}$
$\Rightarrow\dfrac{x^2+2a^2+a^2}{x^2+2a^2}=\dfrac{13}{19}.\dfrac32$
$\Rightarrow1+\dfrac{a^2}{x^2+2a^2}=\dfrac{39}{38}$
$\Rightarrow \dfrac{a^2}{x^2+2a^2}=\dfrac1{38}$
$\Rightarrow x=4a=SC$
