\(\sin R=\dfrac{PQ}{RQ}=\sin60^0=\dfrac{\sqrt{3}}{2}\Leftrightarrow PQ=\dfrac{\sqrt{3}}{2}\cdot6=3\sqrt{3}\)

Áp dụng PTG: \(PR=\sqrt{RQ^2-PQ^2}=\sqrt{36-27}=3\)

Ta có \(\sin\widehat{F}=\dfrac{ED}{EF}=\sin60^0=\dfrac{\sqrt{3}}{2}\Leftrightarrow EF=4\cdot\dfrac{2}{\sqrt{3}}=\dfrac{8\sqrt{3}}{3}\left(cm\right)\\ DF=\sqrt{EF^2-DE^2}=\dfrac{4\sqrt{3}}{3}\left(cm\right)\left(pytago\right)\)

Ta có: ΔABC vuông tại A 

nên \(\widehat{B}+\widehat{C}=90^0\)

hay \(\widehat{C}=30^0\)

Xét ΔABC vuông tại A có 

\(AB=AC\cdot\tan30^0\)

\(\Leftrightarrow AB=\dfrac{4\sqrt{3}}{3}\left(cm\right)\)

hay \(AC=\dfrac{8\sqrt{3}}{3}\left(cm\right)\)

\(\widehat{C}=90-\widehat{B}=90-38=52\)

AC=\(\sin B.BC=\sin38.8\approx4,92cm\)

AB=\(\cos B.BC=\cos38.8\approx6,3cm\)

\(sinB=\dfrac{AC}{BC}\Rightarrow AC=sin60^0.6=3\sqrt{3}\left(cm\right)\)

\(AB=\sqrt{BC^2-AC^2}=\sqrt{6^2-\left(3\sqrt{3}\right)^2}=3\left(cm\right)\)

\(\widehat{C}=90^0-\widehat{B}=90^0-60^0=30^0\)

ΔABC vuông tại A có:
sinB=\(\dfrac{AC}{BC}=\dfrac{AC}{6}\)⇒AC=sin60.6=\(3\sqrt{3}cm\)
cosb=\(\dfrac{AB}{BC}=\dfrac{AB}{6}\)⇒AB=cos60.6=3cm
góc C = 90-góc B=90-30=60 độ

Lời giải:
$\widehat{EMH}=90^0-\widehat{MHE}=90^0-30^0=60^0$

$ME=MH\sin \widehat{MHE}=11.\sin 60^0=\frac{11\sqrt{3}}{2}$ (cm)

$EH=MH\cos \widehat{MHE}=11\cos 60^0=\frac{11}{2}$ (cm)

Bài 1: 

Xét ΔABC vuông tại A có 

\(AB^2+AC^2=BC^2\)

hay \(AB=\sqrt{13}\left(cm\right)\)

Xét ΔABC vuông tại A có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{6}{7}\)

nên \(\widehat{B}=59^0\)

hay \(\widehat{C}=31^0\)

\(\widehat{B}=30^0\)

\(\Leftrightarrow AC=\dfrac{1}{2}\cdot BC\)

hay BC=16cm

\(BC=\dfrac{AC}{cosC}=\dfrac{8}{Cos60}=16cm\)