(a has a direction, counterclockwise is positive, clockwise is negative, and finally the absolute value can be taken. )

Taking the straight line where the vector OA is located as the X axis and the direction of the vector OA as the positive direction of the X axis, constructing a plane rectangular coordinate system obviously includes:

Vector OA = (0, 1), vector OB = (2cos(a), 2 sin(a))

Vector OP = (0, t), vector OQ = (2cos (a) (1-t), 2sin (a) (1-t))

Vector PQ = vector OQ- vector OP = (2cos (a) (1-t), 2sin (a) (1-t)-t)

|PQ| takes the minimum value, that is, the modulus of vector PQ takes the minimum value, and the square of the modulus also takes the minimum value.

| vector pq | 2 = 4cos2 (a) * (1-t) 2+4s in 2 (a) * (1-t) 2-4s in (a) (1-t) t+t.

= 4 (1-t) 2-4t symplectic (a)+4t 2 symplectic (a)+t 2

= 4t^2-8t+4-4t sin(a)+4t^2 sin(a)+t^2

=[5+4 inches (a)] t 2-[8+4 inches (a)] t+4

Think of the above formula as a quadratic function about t;

To get the minimum value of the above formula, t will fall on the symmetry axis of the function.

The symmetry axis of this function is:

t =[8+4 sin(a)]/{ 2[5+4 sin(a)]} =[4+2 sin(a)]/[5+4 sin(a)]= 2-6/[5+4 sin(a)]

That is to say, for a certain included angle a, as long as t satisfies t = 2-6/[5+4sin(a)], then |PQ| takes the minimum value.

According to the meaning of the question, namely:

T0 = 2-6/[5+4 inches (a)]

Because 0

0 & lt2-6/[5+4 inches (a)] < 1/5

10 >30/[5+4 inches (a)] > 9

9 & lt 15+ 12 sin(a)& lt； 10

- 1/2 & lt； Crime (1) & lt-5/ 12

Solve satisfaction-180

Because the included angle is constant, the absolute value of a solved is enough.

Note: the calculation may be wrong, but this idea is not the best or at least correct.