\(\widehat{A}=110^0\)

Chọn đáp án C.

Ta có

 Áp dụng công thức ta có:

V A B C D = 1 6 A B ⇀ . A C ⇀ . A D ⇀ = 1 2

Ta có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

\(\Rightarrow\widehat{C}+\widehat{D}=360^0-\left(\widehat{A}+\widehat{B}\right)=360^0-\left(60^0+100^0\right)=200^0\)

\(\widehat{D}=\dfrac{200^0-40^0}{2}=80^0\)

\(\widehat{C}=\widehat{D}+40^0=80^0+40^0=120^0\)

Ta có tổng các góc trong tứ giác là:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Rightarrow\widehat{C}+\widehat{D}=360^o-\left(\widehat{A}+\widehat{B}\right)=360^o-\left(100^o+60^o\right)\)

\(\Rightarrow\widehat{C}+\widehat{D}=200^o\) (1)

Mà \(\widehat{C}-\widehat{D}=40^o\Rightarrow\widehat{C}=40^o+\widehat{D}\) (2)

Thay (2) vào (1) ta có:

\(40^o+\widehat{D}+\widehat{D}=200^o\)

\(\Rightarrow2\widehat{D}=200^o-40^o\)

\(\Rightarrow2\widehat{D}=160^o\)

\(\Rightarrow\widehat{D}=\dfrac{160^o}{2}=80^o\)

\(\widehat{C}=\widehat{D}+40^o=80^o+40=120^o\)

Chọn B.

A.\(\dfrac{a\sqrt{6}}{3}\)

\(S_{\Delta ACD}=\dfrac{1}{2}AC.AD.sin\widehat{CAD}=\dfrac{a^2\sqrt{3}}{4}\)

\(V=\dfrac{AB.AC.AD}{6}.\sqrt{1+2cos90^0.cos60^0.cos120^0-cos^290^0-cos^260^0-cos^2120^0}=\dfrac{a^3\sqrt{2}}{12}\)

\(\Rightarrow d\left(B;\left(ACD\right)\right)=\dfrac{3V}{S}=\dfrac{a\sqrt{6}}{3}\)

a) 

Mà ta có  (tổng 4 góc trong một tứ giác bằng )

 và 

Góc ngoài của tứ giác tại đỉnh C là 

Xét tứ giác ABCD có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)( tổng các góc trong tứ giác)

\(\Rightarrow\widehat{C}+\widehat{D}=360^0-\widehat{A}-\widehat{B}=360^0-60^0-90^0=210^0\)

Ta có: \(\left\{{}\begin{matrix}\widehat{C}+\widehat{D}=210^0\\\widehat{C}-\widehat{D}=20^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{C}=\left(210^0+20^0\right):2=115^0\\\widehat{D}=\left(210^0-20^0\right):2=95^0\end{matrix}\right.\)