Đáp án: \(\widehat{(HK,CD)}=30^{\circ}\)
Giải thích các bước:
c. Hình chiếu KJ cắt SC tại J
$\to J=(AHK)∩SD$
Ta có: $\displaystyle\left \{ {{KJ\perp SC} \atop {CD\perp (SAC)}} \right.$
$\to\displaystyle\left \{ {{KJ\perp SC} \atop {CD\perp SC}} \right.$
$\to KJ//CD\qquad (1)$
$\Delta ABC$ vuông cân tại A, AH là đường đường nên cũng là đường trung tuyến
$\to AH=HB=HS=\dfrac 12BS=\dfrac{a\sqrt 2}{2}$
Xét $\Delta SAC,$ ta có:
$SC^2=SA^2+AC^2$
$\to SC=\sqrt{SA^2+AC^2}=\sqrt{a^2+2a^2}=a\sqrt 3$
Xét $\Delta SBC,$ ta có:
$\dfrac{SK}{SH}=\dfrac{SB}{SC}$
$\to SK=\dfrac{SH.SB}{SC}=\dfrac{a\sqrt 3}{3}$
$HK=SH.\dfrac{BC}{SC}=\dfrac 1{\sqrt 3}.\dfrac{a\sqrt 2}{2}=\dfrac{a\sqrt 6}{6}$
$\to \dfrac{SK}{SC}=\dfrac{\dfrac{a\sqrt 3}{3}}{a\sqrt{3}}=\dfrac 13$
Xét $\Delta SCD$ có $CD//KJ$
$\to \dfrac{KS}{CD}=\dfrac{SK}{SC}=\dfrac 13=\dfrac{SJ}{SD}$
$\to KJ=\dfrac 13. CD=\dfrac 13. a\sqrt 2=\dfrac{a\sqrt 2}{3}$
Dễ thấy: $DH\perp SB$
$\to \cos\widehat{HSD}=\dfrac{\dfrac{a\sqrt 2}{2}}{a\sqrt 5}=\dfrac{a\sqrt{10}}{5}$
Ta có: $HJ=\sqrt{SJ^2+SH^2-2SJ.SH.\cos\widehat{JSH}}=\sqrt{\bigg(\dfrac{a\sqrt 5}{3}\bigg)^2+\bigg(\dfrac{a\sqrt 2}2\bigg)^2-\dfrac {2.a\sqrt 5}3.\dfrac{a\sqrt 2}2.\dfrac{\sqrt {10}}{10}}=\dfrac{a\sqrt{26}}{6}$
Xét $\Delta HKJ,$ ta có:
$\cos\widehat{HKJ}=\dfrac{HK^2+KJ^2-HJ^2}{2.HK.KJ}$
$\to \cos\widehat{HKJ}=\dfrac{\dfrac 16a^2+\dfrac 29a^2-\dfrac{13}{18}a^2}{2.\dfrac{\sqrt 6}{6}.\dfrac{\sqrt 2}{3}.a^2}=-\dfrac{\sqrt 3}{2}$
$\cos\widehat{(HK,KJ)}=\Big|\cos\widehat{HKJ}\Big|=\dfrac{\sqrt 3}{2}$
$\to \widehat{(HK,KJ)}=30^{\circ}$
Từ $(1)\to \widehat{(HK,KJ)}=\widehat{(HK,CD)}$
$\to \widehat{(HK,KJ)}=\widehat{(HK,CD)}=30^{\circ}$
