Discussion
A layer has a critical frequency of 6 MHz. What is the maximum usable frequency at an incidence whose secant is 3?
Answer: A
Applying the secant law directly:
\[ f_{MUF} = 6 \times 3 = 18 \]
megahertz, that is 18 MHz. A wave at that frequency would escape if sent vertically but is returned at the shallow angle of a long path. This is why the longest hops support the highest frequencies, and why a station may hear a distant contact on a band while nearer stations are inaudible on it.
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