Question

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 1

`\displaystyle (5)/(12) = cot∠B \\ 2(2)/(5) = cot∠A \\ \\ 2(2)/(5) = tan∠B \\ (5)/(12) = tan∠A \\ \\ (5)/(13) = cos∠B \\ (12)/(13) = cos∠A`

Explanation:

`\displaystyle (OPPOSITE)/(HYPOTENUSE) = sin\:θ \\ (ADJACENT)/(HYPOTENUSE) = cos\:θ \\ (OPPOSITE)/(ADJACENT) = tan\:θ \\ (HYPOTENUSE)/(ADJACENT) = sec\:θ \\ (HYPOTENUSE)/(OPPOSITE) = csc\:θ \\ (ADJACENT)/(OPPOSITE) = cot\:θ \\ \\ (10)/(24) = cot∠B → (5)/(12) = cot∠B \\ (24)/(10) = cot∠A → 2(2)/(5) = cot∠A \\ \\ (24)/(10) = tan∠B → 2(2)/(5) = tan∠B \\ (10)/(24) = tan∠A → (5)/(12) = tan∠A \\ \\ (10)/(26) = cos∠B → (5)/(13) = cos∠B \\ (24)/(26) = cos∠A → (12)/(13) = cos∠A`

I am joyous to assist you anytime.