Answer: C
Let \(AC = 3x\) and \(BC = 4x\) in \(\triangle ABC\) \(\therefore AB = \sqrt{AC^2 + BC^2}\) i.e., \(AB = \sqrt{9x^2 + 16x^2}\) i.e., \(AB = \sqrt{25x^2} = 5x\) \(cosec A = \cfrac{AB}{BC} = \cfrac{5x}{4x} = \cfrac{5}{4}\)
Let \(AC = 3x\) and \(BC = 4x\) in \(\triangle ABC\) \(\therefore AB = \sqrt{AC^2 + BC^2}\) i.e., \(AB = \sqrt{9x^2 + 16x^2}\) i.e., \(AB = \sqrt{25x^2} = 5x\) \(cosec A = \cfrac{AB}{BC} = \cfrac{5x}{4x} = \cfrac{5}{4}\)