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If $2 \sin (A + B) = \sqrt{3}$ and $\cos (A - B) = 1$, then find the measures of angles $A$ and $B$. $0 \le A, B, (A + B) \le 90^\circ$.
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$\sin(A + B) = \frac{\sqrt{3}}{2} \Rightarrow A + B = 60^\circ \dots (1)$
$\cos(A - B) = 1 \Rightarrow A - B = 0^\circ \dots (2)$
Solving $(1)$ and $(2)$, we get $A = B = 30^\circ$
$\cos(A - B) = 1 \Rightarrow A - B = 0^\circ \dots (2)$
Solving $(1)$ and $(2)$, we get $A = B = 30^\circ$