As AQ=AR it means ray QR is parallel to the base BC, hence the prism is in the position of minimum deviation ,now μ=sin(A+θ)/2sinA/2 given μ=√3, A=60° therefore √3=sin(60+θ)/2sin60/2 or sin(60+θ)/2=√3×sin30 or sin(60+θ)/2=√3/2=sin60 or (60+θ)=120 or θ=60°