There are two classes of 61 and 62 students in the experimental primary school. If 5 students are transferred from 61 to 62, the number of students in the two classes is equal. If 11 students are transferred from 62 to 61, the number of students in class 61 is twice that of class 62. How many students are there in class 61 and class 62? It must be accurate. Then the number of class 61 is exactly twice that of class 62. How many people are there in class 61 and class 62?

There are two classes of 61 and 62 students in the experimental primary school. If 5 students are transferred from 61 to 62, the number of students in the two classes is equal. If 11 students are transferred from 62 to 61, the number of students in class 61 is twice that of class 62. How many students are there in class 61 and class 62? It must be accurate. Then the number of class 61 is exactly twice that of class 62. How many people are there in class 61 and class 62?

Think like this
If five people are transferred from 61 to 62, the number of people in the two classes will be equal
That means there are 10 more people on June 1 than on June 2
No matter how to transfer the two classes, the total number will remain unchanged
When 11 people are transferred from 62 to class 61, the number of people in class 61 is just twice that of class 62
After the transfer, there are 10 + 11 * 2 = 32 more people in class 61 than in class 62, and the number of people is just twice as many
So after the transfer, there are 32 * 2 = 64 people on June 1 and 32 people on June 2
When there is no adjustment, it is 61 64-11 = 53, 62 32 + 11 = 43
If this problem is junior high school students, using the equation upstairs is also good
What is the value of sin90?
One
One
One
Positive 1 + 1
Radical three / 2
The number of class 61 is equal to that of class 62. The ratio of boys to girls in class 61 is 1:2. The ratio of boys to girls in class 62 is 1:3
Ratio of girls in class 61 to class 62
The ratio of boys to girls in class 61 is 1:2. Assuming that there are x boys in class 61, then there are 2x girls in class 61,
The ratio of boys to girls in class 62 is 1:3. Assuming that there are y boys in class 62, there are 3Y girls in class 62
The number of class 61 is equal to that of class 62, so 3x = 4Y, that is, x = 4 / 3Y
The ratio of the number of girls in class 61 to that in class 62, i.e. 2x / 3Y, is 8:9
How to calculate sin90
Unit circle
In trigonometric functions, there are
sina=y/r
cosa=x/r
tana=y/x
When a = 90 °, y = R
So sin90 ° is Y / r = 1
In a school, the number of students in class 61 is two-thirds of that in class 62. There are eight students in class 62, but the number of students in class 61 is equal to that in class 62. How many students are there in class 62
Forty-eight
There are X students in class 1 and y students in class 2
X/Y=2/3
Y-8=X+8
So x = 32
Y=48
The original number of class 62 was 48
I wish you progress
How much does it cost
sin90=cos0=1
Class 61 has 10 fewer people than class 62. The number of class 1 is 9 / 11 of class 62. How many people are there in class 61 and class 62?
Class 6-2 10 / (1-9 / 11)
＝10/（2/11）
= 55
Class 61 55 * 9 / 11 = 45
Class 61 45 class 62 55
61 is nine out of eleven of 62
So there are 9 copies of 61, 11 copies of 62, because 61 has 10 fewer people than 62, so 11-9 = 2, 10 people are two of them
One share is 5 people 61, 9 times 5 = 45 people 62, 11 times 5 = 55 people
Sin0 degrees equals
How much is it
Sin0 equals 0; sin30 equals 1 / 2; sin60 equals root 3 / 2; sin90 equals 1
Cos 0 is equal to 1; Cos 30 is equal to root 3 / 2; Cos 60 is equal to 1 / 2; Cos 0 is equal to 0
Class 61 is 45 times more than class 62, that is, class 62 is 45 times less than class 61
FALSE
(wrong)
Correct answer:
1÷(1+1/45)=1/46
The number of class 62 is 1 / 46 less than that of class 61
How to find sin0 °, sin90 ° and sin180,
You can draw the unit circle, otherwise use the trigonometric function formula of high school
sin0°=sin（45°-45°）=sin45°cos45°-cos45°sin45°=0
sin90°=sin（45°+45°）=sin45°cos45°+cos45°sin45°=2×√2/2×√2/2=1
Sin180 ° = sin (90 ° + 90 °) = sin90 ° cos90 ° + cos90 ° sin90 ° = 0 〔 (sin ^ 90 °) ^ 2 + (cos90 °) ^ 2 = 1, then cos90 ° = 0 〕