The sum of the moments about point A is zero, so we have:
$150(2a) + 150(2a+6) - R_B(4a+6) = 0$
Solving for $R_B$, we get:
$R_B = \frac{150(4a+6)}{4a+6}$
Since the sum of the vertical forces is also zero, we have:
$R_A + R_B = 150 + 150$
Substituting $R_B$
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