(1) The coordinate of point B is (5,5 √ 3), and AB= 10.

We can know that sin ∠ bao = 5 ∠ 3/ 10 = ∠ 3/2.

So ∠BAO is an angle of 60 degrees (and CA vertical OA can be obtained according to the meaning of the question)

(2) area △ OPQ S= bottom OQ* height * 1/2.

Suppose the speed of P and Q is V (the speed of two points is the same, which is the meaning in the question).

Here we can see that the bottom OQ = vt+2; The height is the abscissa of point P. If the vertical line OA of PE intersects with point E, it can be seen that the height = 10 (this is the abscissa of point A) -AE= 10- 1/2 AP (according to the 60-degree angle result obtained earlier) = 10-vt/2.

Therefore, the area of △OPQ is s = (vt+2) (10-vt/2)/2-★.

According to Figure 2, when t=5 and s=30, we can calculate v=2 or v= 1.6 by solving the above formula.

At this time, the most difficult point of this essay comes-parabola S=(vt+2)( 10-vt/2)/2. After simplification, the symmetry axis (t=-b/2a) is t = 9/v. From Figure ②, we can see the symmetry axis of t< parabola T; 5, so v= 1.6 does not meet the problem.

So v=2 is the only solution to this small problem.

(3) The formula marked ★ in the above question is the functional relationship between area s and time t.

At the same time, when the parabola is at the vertex, the value of S is the largest. At this time t=9/v=4.5.

From the analysis in (2) above, it can be seen that the abscissa of point P is = 10-vt/2, and after substitution, the abscissa is =5.5 and the ordinate is =( 10-5.5)√3 =4.5√3.

That is, point P(5.5, 4.5√3)

(4) This short answer is difficult. If the whole question is 15, then this quiz should not exceed 3 points. You can consider giving up during the exam.

The solutions that can be scored are as follows:

When P moves to point B, it can be calculated that the coordinates of point P are (5,5 √ 3), the coordinates of point Q are (0, 12) and the angle POQ is 30 degrees.

At this point, the respective lengths of the three sides of the triangle can be calculated.

We find that the square of QO >: the square of PO+the square of PQ. So the OPQ angle is obtuse (derivative of cosine theorem)

When the point P moves along AB, the size of ∠OPQ increases with the increase of time t .. So there must be a point that makes the angle OPQ=90 degrees when P moves from A to B.

Similarly, it can be assumed that P moves to C, and at this time, according to the derivation of cosine theorem, we can get what angle OPQ is. According to the movement of P along BC, the size of ∠OPQ decreases with the increase of time t. If P is in C and the angle OPQ is obtuse, there will definitely be no right angle in the middle. Whether the OPQ angle is acute or right. Then there must be a point p between BC, so the OPQ angle is a right angle.

For the above reasons, the final actual answer is that there are two such points p.