Correct Answer - C
From free body diagrams
`Nsin alpha=Ma`…(i)
`N+mA sin alpha=mgcos alpha` ….(ii)
`mg sin alpha+mA cos alpha=ma` ….(iii)
On solving equations (i), (ii) and (iii)
`A=(mg cos alpha sin alpha)/(M+m sin^(2)alpha), a=((M+m)g sin alpha)/(M+msin^(2)alpha)`
Now, `vec(a)_("block")=vec(a)_("block/wedge")+vec(a)_("wedge")`
`:. vec(a)_("block")=a cosalphahat(i)-a sin alphahat(j)-Ahat(i)`
`(a cos alpha-A)hat(i)-a sin alphahat(j)`
`:. tanbeta=(a sin alpha)/(a cos alpha-A)=(1+m/M)tanalpha`