EASA NAV(A)advanced
23. Given WCA −12°, TH 125°, MC 139°, DEV 2°E, find TC, MH, and CH.
Answer
TC 137°, MH 127°, CH 125°
Explanation
Using $TC=TH-WCA$, TC is 137°. With 12°E variation, MC 139° corresponds to MH 127°; applying 2°E deviation gives CH 125° under the bank's sign convention. **Key takeaway:** Apply WCA, variation, and deviation in sequence. **Glossary:** WCA: wind correction angle; DEV: deviation.
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