1. **State the problem:** Given that $\sec 2A = \csc (A - 27^\circ)$ and $2A$ is an acute angle, find the measure of $\angle A$. 2. **Recall the definitions:** - $\sec \theta = \frac{1}{\cos \theta}$ - $\csc \theta = \frac{1}{\sin \theta}$ 3. **Rewrite the equation using these definitions:** $$\frac{1}{\cos 2A} = \frac{1}{\sin (A - 27^\circ)}$$ 4. **Cross-multiply to get:** $$\sin (A - 27^\circ) = \cos 2A$$ 5. **Use the identity $\cos 2A = \sin (90^\circ - 2A)$:** $$\sin (A - 27^\circ) = \sin (90^\circ - 2A)$$ 6. **Since $\sin x = \sin y$, then either:** - $A - 27^\circ = 90^\circ - 2A + 360^\circ k$ or - $A - 27^\circ = 180^\circ - (90^\circ - 2A) + 360^\circ k$ where $k$ is any integer. 7. **Solve the first equation:** $$A - 27 = 90 - 2A + 360k$$ $$A + 2A = 90 + 27 + 360k$$ $$3A = 117 + 360k$$ $$A = \frac{117 + 360k}{3} = 39 + 120k$$ 8. **Solve the second equation:** $$A - 27 = 180 - 90 + 2A + 360k$$ $$A - 27 = 90 + 2A + 360k$$ $$A - 2A = 90 + 27 + 360k$$ $$-A = 117 + 360k$$ $$A = -117 - 360k$$ 9. **Check for $A$ such that $2A$ is acute (i.e., $0^\circ < 2A < 90^\circ$):** - From first solution with $k=0$: $A=39^\circ$, so $2A=78^\circ$ (acute) - Other values for $k$ will be outside the acute range. - From second solution, $A$ is negative or too large, discard. 10. **Final answer:** $$\boxed{39^\circ}$$