a: \(\widehat{C}=60^0\)

\(AC=6\sqrt{3}\left(cm\right)\)

\(BC=12\sqrt{3}\left(cm\right)\)

:< giải hộ mình với ~

a ,   Δ A B C ,   A ⏜ = 90 0 , A H ⊥ B C g t ⇒ A H = B H . C H = 4.9 = 6 c m Δ A B H ,   H ⏜ = 90 0   g t ⇒ tan B = A H B H = 6 4 ⇒ B ⏜ ≈ 56 , 3 0 b ,   Δ A B C ,   A ⏜ = 90 0 , M B = M C g t ⇒ A M = 1 2 B C = 1 2 .13 = 6 , 5 c m S Δ A H M = 1 2 M H . A H = 1 2 .2 , 5.6 = 7 , 5 c m 2

a)xét \(\Delta\)ABC vuông tại A có

\(\widehat{B}+\widehat{C}=90'\Rightarrow\widehat{C}=90'-30'=60'\)

\(\sin C=\frac{AB}{BC}\Rightarrow BC=\frac{AB}{\sin B}=\frac{6}{\sin30'}=12\left(cm\right)\)

\(\tan B=\frac{AC}{AB}\Rightarrow AC=AB.\tan B=6.\tan30'=2\sqrt{3}\left(cm\right)\)

b)Xét \(\Delta ABC\left(\widehat{BAC}=90'\right)AHvuôngócBC\)

\(AB^2=BC.HB\Rightarrow HB=\frac{AB^2}{BC}=\frac{6^2}{12}=3cm\)

\(AH.BC=AB.AC\)

\(\Rightarrow AH=\frac{AB.AC}{BC}=6.2\sqrt{3}=12\sqrt{3}cm\)(1)

VÌ AM LÀ ĐƯỜNG TRUNG TUYẾN CỦA TG ABC NÊN

\(MB=MC=\frac{BC}{2}=\frac{12}{2}=6cm\)

MÀ\(MB=MH+HB\)

\(\Rightarrow MH=MB-HB=6-3=3cm\)(2)

TỪ (1)và (2) SUY RA

\(S\Delta AHM=\frac{1}{2}AH.HM=\frac{1}{2}.12\sqrt{3}.3=18\sqrt{3}\approx31.18\left(cm^2\right)\left(do\Delta AHMvuôngtạiH\right)\)

\(a,BC=\sqrt{AB^2+AC^2}=13\left(cm\right)\\ HTL:\left\{{}\begin{matrix}AH=\dfrac{AB\cdot AC}{BC}=\dfrac{60}{13}\left(cm\right)\\BH=\dfrac{AB^2}{BC}=\dfrac{25}{13}\left(cm\right)\end{matrix}\right.\\ b,AM=\dfrac{1}{2}BC=\dfrac{13}{2}\left(cm\right)\left(trung.tuyến.ứng.cạnh.huyền\right)\\ \Rightarrow HM=\sqrt{AM^2-AH^2}=\dfrac{119}{26}\left(cm\right)\\ \Rightarrow S_{AHM}=\dfrac{1}{2}AH\cdot HM=\dfrac{1}{2}\cdot\dfrac{60}{13}\cdot\dfrac{119}{26}=\dfrac{1785}{169}\left(cm^2\right)\)