(3)当射线OP ,OQ 旋转到同一条直线上时,有以下3种情况,
①如图
此时180POA AOB BOQ ︒
∠+∠+∠=,即206010180t t ︒︒︒︒++=,解得4t =; ②如图
此时点P 和点Q 重合,可得360POA AOB BOQ ︒
∠+∠+∠=,即206010360t t ︒︒︒︒++=,解得10t =;
③如图
此时180BOQ BOP ︒∠+∠=,即1060(36020)180t t ︒︒︒︒︒
⎡⎤+--=⎣⎦,解得16t =, 综合上述,4t =或10t =或16t =;
(4)由题意运动停止时3602018t ︒︒=÷=,所以018t <≤,
①当04t <<时,如图,
此时OA 为POQ ∠的“二倍角线”,2AOQ POA ∠=∠,
即6010220t t ︒︒︒+=⨯,解得2t =;
②当410t ≤<时,如图,
