bài này dài lắm ko ai giải đâu

dai den bao gio moi xong lol

Đặt AB = x>0 , AC = y>0 , BC = z>0

Theo định lí Cosin, ta có : \(x^2=y^2+z^2-2yz.cos45^o\Leftrightarrow64=96+z^2-8\sqrt{3}z\)\(\Leftrightarrow\left[\begin{array}{nghiempt}z=4+4\sqrt{3}\\z=-4+4\sqrt{3}\end{array}\right.\)

Vậy BC = \(4+4\sqrt{3}\) hoặc BC = \(4\sqrt{3}-4\)

+Với \(\begin{cases}x=8\\y=4\sqrt{6}\\z=4+4\sqrt{3}\end{cases}\) thì \(cosB=\frac{1}{2}\Rightarrow\widehat{B}=60^o\)

+Với \(\begin{cases}x=8\\y=4\sqrt{6}\\z=4\sqrt{3}-4\end{cases}\) thì \(cosB=-\frac{1}{2}\Rightarrow\widehat{B}=120^o\)

Chứng minh như sau : Kẻ đường cao AK (K thuộc BC)

Trong tam giác vuông AKC có : \(AK=sinC.AC\)

Ta có :  \(S_{\Delta ABC}=\frac{1}{2}BC.AK=\frac{1}{2}BC.AC.SinC\)

+Với \(\begin{cases}x=8\\y=4\sqrt{6}\\z=4+4\sqrt{3}\end{cases}\) thì  \(S_{\Delta ABC}=\frac{1}{2}AC.BC.sin45^o=\frac{1}{2\sqrt{2}}.4\sqrt{6}.\left(4+4\sqrt{3}\right)=24+8\sqrt{3}\)

+Với \(\begin{cases}x=8\\y=4\sqrt{6}\\z=4\sqrt{3}-4\end{cases}\) thì \(S_{\Delta ABC}=\frac{1}{2}AC.BC.sin45^o=\frac{1}{2\sqrt{2}}.4\sqrt{6}.\left(-4+4\sqrt{3}\right)=24-8\sqrt{3}\)

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) Ta có: 

\(\widehat{A}=180^o-60^o-45^o=75^o\)

Áp dụng định lý sin ta có:

\(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)

\(\Rightarrow AC=\dfrac{BC\cdot sinB}{sinA}\)

\(\Rightarrow AC=\dfrac{a\cdot sin60^o}{sin75^o}=a\cdot\dfrac{3\sqrt{2}-\sqrt{6}}{2}\) 

\(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\)

\(\Rightarrow AB=\dfrac{BC\cdot sinC}{sinA}\)

\(\Rightarrow AB=\dfrac{a\cdot sin45^o}{sin75^o}=a\cdot\left(\sqrt{3}-1\right)\) 

b) \(cos75^o\)

\(=cos\left(30^o+45^o\right)\)

\(=cos30^o\cdot cos45^o-sin30^o\cdot sin45^o\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{\sqrt{2}}{2}\cdot\left(\dfrac{\sqrt{3}-1}{2}\right)\)

\(=\dfrac{\sqrt{6}-\sqrt{2}}{4}\left(dpcm\right)\)