The clock runs from 12 o clock to 3 o clock, and the hour and minute hands form several right angles

From 12 o'clock to 3 o'clock, the minute hand goes 3 times, turns 1080 degrees, and the hour hand circle, turns 90 degrees, and the angle between the two changes continuously from 0 degrees to 990 degrees, of which 90, 270, 450, 630, 810, 990 are right angles.

Including the three-point one, there are six right angles.

Two more times at 12 o'clock, two more times at one point, once at two o'clock, once at 3 o'clock, a total of 6 times.

Because the hour hand moves relatively slowly, we can assume that the hour hand does not move for each unit hour (only 30 degrees in an hour), then the minute hand can make a circle of 360 degrees (330 degrees) and form two right angles to a fixed straight line in a circle. So, 12 to 3 can be 6 right angles. Draw your own picture and see it.

The easiest way:

For every 1 hour of the hour hand, the minute hand makes one turn (306 degrees).

In a 306-degree circle (your clock dial), any radius (hour hand) has only one diameter (minute hand) perpendicular to it within this circle.

So after every hour that has passed, the minute hand will sweep across the dial, and it will inevitably sweep over the diameter perpendicular to the hour hand. The diameter is two radii, so it's two.

Therefore, for every hour that passes, the hour and minute hands must be perpendicular twice.

Which little alarm clock to experiment with?

Agree with the two comrades upstairs, the Chinese are just dead studying.

The problem is definitely faster than the math, and it is intuitive.

Why can the developed countries of the West develop rapidly, because they do not waste time on such boring problems.

Practice is the only criterion for testing the truth, just take a table and experiment to find out.

2*30+90) (Another 3 11 points are missing.)

Well! Right! At 2:27 and 3/11, the hour and minute hands are rubberized at right angles.

At 12 o'clock, the angle between the two pins is 0°

At 12:30, the hour hand turns to 360*(30 12*60)=15°, and the minute hand turns 360*(30 60)=180°, so the angle of hail between the two is 180-15=165°

Answer]: B Reference Analysis]: It can be seen from the thick bush of the question that within an hour, the minute hand turns once, so there are only 2 stool grips for the cherry blossom lift that can form a right angle to the hour hand, so B is selected for this question.

At 12 o'clock, the angle between the two needles is 0°

At 12:30, the hour hand turns 360 * (30 12 * 60) = 15 °, the minute hand turns 360 * (30 60) = 180 °, so the big silver angle of the two is 180-15 = 165 °, 7, 165 °, 1, the minute hand turns 5 30 = 150 degrees.

Turn the hour hand 30 degrees.

15+150=165 degrees,1,180-15=165°,0,

Total 220 degrees. Therefore, when the minute hand and the hour hand coincide, the angle of the minute hand = the angle of the Hongling taken by the hour hand + 220 degrees] When the clock has traveled for 40 minutes, the old hour hand and the minute hand coincide. When set to y minutes, a right angle is formed.

Or when the clock runs the minute or minute, the angle between the hour and minute hands is 90 degrees. It is a flat angle when it is set to z minutes. When the minutes have elapsed, the angles between the hour and minute hands are flat.

Note: The minute hand walks for 1 minute, the angle of walking is 360 60 = 6 degrees, and the hour hand goes 360 12 60 = degrees.

The hour hand is at right angles to the minute hand, and the angle difference between the instant hand and the minute hand is 90 degrees or 270 degrees, which should theoretically be 2 times, and it should be verified:

According to the angle difference speed difference = number of minutes, 90 and 4 11 60 can be obtained, which means that after 16 and 4 11 minutes, the hour hand and the minute hand are perpendicular for the first time;In the same way, 270 and 1 11 60 means that after 49 and 1 11 minutes, the hour hand is perpendicular to the minute hand for the second time.

There are two times. The minute hand rotates once in an hour, and there is a chance to be at right angles on the left and right sides of the clock, as shown in the figure below, and the times are as follows:

From 12 o'clock to 13 o'clock, after 60 minutes, 15 (1-1 12).

16 (4 11) (min).

49 (1 11) (min).

So from 12 o'clock to 13 o'clock, there are two chances that the hour hand and the minute hand can be at right angles, at 12:16 (4 11) and 12:49 (1 11).

From 12 o'clock to 13 o'clock, how many times does the clock hand be at right angles to the minute hand?

2 times, respectively:

16 and 4 11 points.

49 and 1 11 points.

At 12:16 and 4 11 and 12:49 and 1 11, the hour and minute hands of the clock can be at right angles.

To know at which moment vertically.

First of all, it is necessary to understand the vertical post.

The interval between the hour and minute hands.

Because it's 90 degrees.

So draw conclusions.

In a 60-cell disc, they are spaced 15 squares apart.

Thus derived. The range of the hour hand is between the 35th and 40th divisions.

The minute hand is between 0 and 60 stops.

And because they're spaced 15 squares, there are two outcomes.

The first result is that the minute hand scale - the hour hand scale = 15

The second result is that the hour hand scale - the minute hand scale = 15

You can list a system of equations: Let the hour hand scale be x, and the minute hand scale be y|x-y|=15

x-35）/5 = y/60

350 First Result x=405 11 y=240 11Second Result x=435 11 y=600 11 So the final standard answer is.

The first result is 7:21 and 9 11.

The second result is 7:54 and 6 11 points.

There are two moments when the hour and minute hands are at right angles. The hour hand travels by the clock every minute, the minute hand travels by 6° per minute, and the hour and minute hands are set at right angles after going for x minutes. (1) 210+ solution x = 240 11 That is, at 7:240 11 minutes, the hour and minute hands are at right angles.

2) Solution x=600 11 i.e. at 7:600 11 minutes, the hour and minute hands are again at right angles.

Let the circumference of the clock be LThe length of the arc traveled in x minutes is lx 60The length of the arc traveled by the hour hand is 1 12

The arc length before the seven-point position is 7l 12So there is 7l 12+1 12*lx 60-lx 60=l 4The solution gives 1 3 = 11 12 * x 60 and x 240 11, which is 7:22

at right angles (30, 7-90) (

21 and 9 11 points.

That is, 7:21 and 9 11.

or (30 7+90) (.)

54 and 6 11 points.

i.e. 7:54 and 6 11.

Because the scale on the clock divides a circle into 12 equal parts, each of which is 30°, with the help of a graph, find out the difference between the hour and minute hands at 8:30 and multiply it by 30°

Solution: At 8:30, there is a big difference between the hour hand and the minute hand There are 12 numbers in the clock, and the angle between each adjacent two numbers is 30°, and the angle between the minute hand and the hour hand at 8:30 is .

So the answer is: 75

Because at 8:30, the hour hand points between 8 and 9, and the minute hand points to 6 The 12 numerals of the clock, and the angle between each two adjacent numbers is 30°, so the angle between the hour and minute hands and the hour hand at 8:30 is 2 30° + 15° = 75 degrees

So the answer is: 75