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If $\sin(A + 2B) = \frac{\sqrt{3}}{2}$ and $\cos(A+ 4B) = 0$, $A > B$, find $A$ and $B$.
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(b) $\sin(A + 2B) = \frac{\sqrt{3}}{2} \Rightarrow A + 2B = 60^\circ$ (1/2 Mark)
$\cos(A + 4B) = 0 \Rightarrow A + 4B = 90^\circ$ (1/2 Mark)
Solving to get $A = 30^\circ$, $B = 15^\circ$ (1 Mark)
(b) $\sin(A + 2B) = \frac{\sqrt{3}}{2} \Rightarrow A + 2B = 60^\circ$ (1/2 Mark)
$\cos(A + 4B) = 0 \Rightarrow A + 4B = 90^\circ$ (1/2 Mark)
Solving to get $A = 30^\circ$, $B = 15^\circ$ (1 Mark)