Bài 6:
a.
$AH=\sqrt{AB^2-BH^2}=\sqrt{13^2-5^2}=12$ (cm)
$\sin B=\sin \widehat{ABH}=\frac{AH}{AB}=\frac{12}{13}$
$\sin C=\sin \widehat{BAH}=\frac{BH}{AB}=\frac{5}{13}$
b.
Áp dụng HTL trong tam giác vuông:
$AH^2=BH.CH=3.4=12$
$\Rightarrow AH=\sqrt{12}=2\sqrt{3}$
$AB=\sqrt{BH^2+AH^2}=\sqrt{3^2+12}=\sqrt{21}$ (theo định lý Pitago)
$\sin B=\frac{AH}{AB}=\frac{2\sqrt{3}}{\sqrt{21}}$
$\sin C=\frac{BH}{AB}=\frac{3}{\sqrt{21}}$