Find acute angles `Aa n dB ,` if `sin(A+2B)=(sqrt(3))/2a n dcos(A+4B)=0,A > Bdot`

Explanation

To find the acute angles \( A \) and \( B \) given the equations \( \sin(A + 2B) = \frac{\sqrt{3}}{2} \) and \( \cos(A + 4B) = 0 \), we can follow these steps:

### Step 1: Analyze the first equation
We know that \( \sin(A + 2B) = \frac{\sqrt{3}}{2} \). The sine of an angle is \( \frac{\sqrt{3}}{2} \) at \( 60^\circ \) (or \( \frac{\pi}{3} \) radians). Therefore, we can set up the equation:
\[ A + 2B = 60^\circ \quad \text{(1)} \]

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