Đáp án:

Gốc tọa độ `O(0;0;0)=>SH bot (ABC)`

Theo đề bài ta có: `Oz bot SH`

Ta có: `B in Ox=>B(60;0;0)`

`@ A in Oy=>A(0;60sqrt3;0)`

`@C (-60;0;0)`

Do `OA=(asqrt3)/2`    `OH=(asqrt3)/6`

`a,` Đ

`b,` Ta có:

`vec{SB}=((0-60);(0-20sqrt3);(0-110))=(60;-20sqrt3;-110)`

`vec{AB}=((60-0);(0-60sqrt3);0)=(60;-60sqrt3;0)`

`->` Sai. 

`c,` Ta có: `vec{n}=[vec{SB};vec{AB}]=(11sqrt3;11;4sqrt3)`

`->` Sai.

`d,` Ta có: `vec{m}=[vec{SC};vec{AC}]=(11sqrt3;-11;-4sqrt3)`

`@cos(vec{m};vec{n})=|11sqrt3.11sqrt3-11.11-4sqrt3.4sqrt3|/(sqrt{(11sqrt3)^2+11^2+(4sqrt3)^2}.sqrt{(11sqrt3)^2+(-11)^2+(4sqrt3)^2})``~~0;36`

`->` Sai.