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Along the normal both $$A$$ and $$B$$ have same acceleration.

B will are along the Inclined path

$$a_{A} \cos(90^o - \theta) = a_{B} cos \theta $$

$$a_{A} \sin \theta = a_{B} cos \theta$$

$$a_{A} = \dfrac{a}{sin \theta }(cos \theta) $$

$$ = a \cot \theta$$

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