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If $\sin(2A + 3B) = 1$ and $\cos(2A - 3B) = \frac{\sqrt{3}}{2}$, $0^{\circ} < 2A + 3B \leq 90^{\circ}$, A $>$ B, then find A and B.
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$\sin (2A + 3B) = 1 \Rightarrow 2A + 3B = 90^{\circ}$ --- (1)
$\cos (2A - 3B) = \frac{\sqrt{3}}{2} \Rightarrow 2A - 3B = 30^{\circ}$ --- (2)
Solving (1) and (2), we get
A = $30^{\circ}$ and B = $10^{\circ}$
$\cos (2A - 3B) = \frac{\sqrt{3}}{2} \Rightarrow 2A - 3B = 30^{\circ}$ --- (2)
Solving (1) and (2), we get
A = $30^{\circ}$ and B = $10^{\circ}$