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In \(\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \mathrm{BC}=5 \ \mathrm{cm}, \mathrm{AC}-\mathrm{AB}=1 \mathrm{~cm}\), then \(\frac{1+\sin (\mathrm{C})}{1+\cos (\mathrm{C})}\) is

Question and options

In △ABC,∠B=90∘,BC=5 cm,AC−AB=1 cm\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}, \mathrm{BC}=5 \ \mathrm{cm}, \mathrm{AC}-\mathrm{AB}=1 \mathrm{~cm}, then 1+sin⁡(C)1+cos⁡(C)\frac{1+\sin (\mathrm{C})}{1+\cos (\mathrm{C})} is
  1. 56\frac{5}{6}
  2. 65\frac{6}{5}
  3. 22
  4. 2518\frac{25}{18}

Answer and explanation

Correct answer: D - \(\frac{25}{18}\)
AC^2 - AB^2 = BC^2 = 25 and AC-AB = 1 give AC+AB = 25, so AC = 13, AB = 12. sin C = AB/AC = 12/13, cos C = 5/13. (1+12/13)/(1+5/13) = 25/18.

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