In the arrangement shown in figure , if the acceleration of B is $\vec{a}$ , then find the acceleration of A 

Along the normal both $A$ and $B$ have same acceleration.
B will are along the Inclined path 
 $a_A\cos (90^o-\theta )=a_{B}cos\theta$ 
$a_A\sin \theta =a_{B}cos\theta$ 
$a_A=\frac{a}{sin\theta }(cos\theta )$
$=a\cot \theta$