求三角型两边长度
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AB^2 + BC^2 = AC^2 设 AB = 4k
BC = 3k
(4k)^2 + (3k)^2 = 529^2 16k^2 + 9k^2 = 529^2 25k^2 = 529^2 k^2 = (529^2) / 25 k = 529/5 AB = 4*529/5 = 423.2 cm BC = 3*529/5 = 317.4 cm
tan(a)=3/4 角a=36.87"c cos(a)=ab/ac cos(36.87)=ab/529 ab=423.2cm sin(a)=bc/ac sin(36.87)=bc/529 bc=317.4cm
BC = 3k
(4k)^2 + (3k)^2 = 529^2 16k^2 + 9k^2 = 529^2 25k^2 = 529^2 k^2 = (529^2) / 25 k = 529/5 AB = 4*529/5 = 423.2 cm BC = 3*529/5 = 317.4 cm
tan(a)=3/4 角a=36.87"c cos(a)=ab/ac cos(36.87)=ab/529 ab=423.2cm sin(a)=bc/ac sin(36.87)=bc/529 bc=317.4cm
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