Find the cotangent, cosine, and tangent of both angles A and B.If you could put it in this format: Cotangent - Cosine - Tangent - that would be epic :^D

Answer:[tex]\displaystyle \frac{5}{12} = cot∠B \\ 2\frac{2}{5} = cot∠A \\ \\ 2\frac{2}{5} = tan∠B \\ \frac{5}{12} = tan∠A \\ \\ \frac{5}{13} = cos∠B \\ \frac{12}{13} = cos∠A[/tex]Step-by-step explanation:[tex]\displaystyle \frac{OPPOSITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOSITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOSITE} = csc\:θ \\ \frac{ADJACENT}{OPPOSITE} = cot\:θ \\ \\ \frac{10}{24} = cot∠B → \frac{5}{12} = cot∠B \\ \frac{24}{10} = cot∠A → 2\frac{2}{5} = cot∠A \\ \\ \frac{24}{10} = tan∠B → 2\frac{2}{5} = tan∠B \\ \frac{10}{24} = tan∠A → \frac{5}{12} = tan∠A \\ \\ \frac{10}{26} = cos∠B → \frac{5}{13} = cos∠B \\ \frac{24}{26} = cos∠A → \frac{12}{13} = cos∠A[/tex]I am joyous to assist you anytime.