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 $(1)\to \widehat{(HK,KJ)}=\widehat{(HK,CD)}$

$\to \widehat{(HK,KJ)}=\widehat{(HK,CD)}=30^{\circ}$

 

Bạn tham khảo ạ, mình vẽ hình hơi sai xíu mong bạn thông cảm :(