a: góc C=60 độ
Xét ΔABC vuông tại A có sin B=AC/BC
nên AC/BC=1/2
=>AC=15(cm)
=>AB=15 căn 3(cm)
b: \(AH=\dfrac{15\cdot15\sqrt{3}}{30}=\dfrac{9}{2}\sqrt{3}\left(cm\right)\)
\(BH=\dfrac{AB^2}{BC}=225\cdot\dfrac{3}{30}=22.5\left(cm\right)\)
c: \(\dfrac{BD}{EC}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}\)
\(=\left(\dfrac{BH}{CH}\right)^2\cdot\dfrac{AC}{AB}=\left(\dfrac{AB}{AC}\right)^4\cdot\dfrac{AC}{AB}=\left(\dfrac{AB}{AC}\right)^3\)