Step-by-step explanation:
[tex]sin K = \frac{JH}{KH} ...(1) \\ \\ cos H= \frac{JH}{KH} ...(2) \\ \\ from \: equations \: (1) \: and \: (2) \: it \: is \: \\ obvious \: that: \\ \\ \huge \boxed{\red{ \sin K} \purple{= } \orange{\cos H }}\\ [/tex]
Thus, sin K is the correct answer.
