Câu hỏi của Hồ Việt Hoàng

a: BC=AD=5a=>AB=13asin B=AC/AB=12/13; cos B=BC/AB=5/13\(\dfrac{sinB-cosB}{sinB+cosB}=\dfrac{\dfrac{12}{13}-\dfrac{5}{13}}{\dfrac{12}{13}+\dfrac{5}{13}}=\dfrac{7}{17}\)b: Kẻ CH vuông góc AB=>CH=CA*CB/AB=60a^2/13a=60a/13Câu trả lời: a: BC=AD=5a=>AB=13asin B=AC/AB=12/13; cos B=BC/AB=5/13\(\dfrac{sinB-cosB}{sinB+cosB}=\dfrac{\dfrac{12}{13}-\dfrac{5}{13}}{\dfrac{12}{13}+\dfrac{5}{13}}=\dfrac{7}{17}\)b: Kẻ CH vuông góc AB=>CH=CA*CB/AB=60a^2/13a=60a/13