Câu trả lời:
Ta có: \(\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow\dfrac{AB}{3}=\dfrac{AC}{4}\Rightarrow\dfrac{AB^2}{9}=\dfrac{AC^2}{16}\)
Ta có tam giác ABC vuông tại A\(\Rightarrow AB^2+AC^2=BC^2\)
Hay \(AB^2+AC^2=15^2=225\)
Ta có \(\dfrac{AB^2}{9}=\dfrac{AC^2}{16}=\dfrac{AB^2+AC^2}{9+16}=\dfrac{225}{25}=9\)
Do đó \(\dfrac{AB^2}{9}=9\Rightarrow AB^2=9\cdot9=81\Rightarrow AB=9\)
\(\dfrac{AC^2}{16}=9\Rightarrow AC^2=9\cdot16=144\Rightarrow AC=12\)