Knowing that in ABC, AB 1, BC 2, then what is the range of values for C? Seek the process.

First of all, c is greater than 0, bc is greater than ab, so c is less than 90. Then from the sinusoidal theorem, we know that sinc = 1 2sina, and sina is less than or equal to 1, so sinc is less than 1 2, and c should be less than or equal to 30 or greater than or equal to 150, and from c less than 90, c is known to be greater than 0 and less than or equal to 30.

Alternatively, you can use the method of drawing and see the answer directly. Make BC, then take B as the center of the circle, 1 as the radius to make a circle, select any point on the circle is A, connect AC, let A rotate on the circle, you can see the change range of angle C.

Wish! I especially hope you can learn the latter method, which is very simple.

Good luck with your studies! (Best wishes as a prospective university student of Zhejiang University).

Because c=ab=1, a=bc=2, b=ac

According to the sum of the two sides is greater than the third side, and the difference between the two sides is less than the third side, 1 b 3 can be known, according to the cosine theorem.

cosc＝(a²＋b²－c²)／2ab

4＋b²－1）／4b

3＋b²）／4b

3／4b＋b／4

2√3/4b*b/4

So 0

ab/sinc=bc/sina

1/sinc=2/sina

sinc=sina/2∈（0,1/2]

bc>ab

So a>c

c is an acute angle. c∈(0°,30°]

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This seems to be in the textbook.

2a+2>22-8=142a>14-2=12

a>6 and 2a + 2 < 22 + 8 = 30

2a<30-2=28

a<28÷2=14

So the rule of 6 division:

Where the dividend contains a divisor, 6 times, the period method is: the dividend contains 4 times the quotient: half of the previous digit is added, and the base digit is subtracted once.

The dividend contains 5 times the quotient: the previous digit is added to the complement by half, and the base does not move.

The dividend contains 6 times the quotient: half of the complement is added to the previous position, and the complement is added once to the base.

Example: 35568 + 78 = 456 (the complement of 78 is 22) Calculation sequence: 355 contains 4 times the divisor, so the front plus 11, the base minus 22, the middle contains the divisor 5 times, the front plus 11, the base does not move, and 45-468.

468 contains 6 times the divisor, add 11 to the front, add 22 to the standard, and get 456 (quotient).

ab=6,ac=3, what is the value range of bc?

Assuming that BC is the shortest, that is, BC is perturbed with AB heavy potato Hu, and ABC is in a line (AB BC=AC), then the shortest BC is several denier AB-AC=3

Assuming that BC is the longest, that is, AC is an extension of AB, then BC=AB AC=6 3=9 so the BC range should be: 9>BC>3

Triangle: The sum of the two sides is greater than the third side, the difference between the two sides is less than the third side, and the balance of the defeat is ab+ac=9 > bc

So ab - ac = 3 so the range of the c's c's side value is 3 to 9, i.e. (3,9).

ab=6,ac=3, what is the value range of bc?

Assuming that BC is the shortest, that is, BC is perturbed with AB heavy potato Hu, and ABC is in a line (AB BC=AC), then the shortest BC is several denier AB-AC=3

Assuming that BC is the longest, that is, AC is an extension of AB, then BC=AB AC=6 3=9 so the BC range should be: 9>BC>3

Summary. 1 a 3, -1 b 1, -1 b 1, -1+1 a-b 1+3, i.e., -1 a-b 4, so the range of values of a-b is.

If 1 a 3, -1 b<1, the value of ab is in the range of .

Need detailed process thank you.

Hello, I am glad to answer for you, if 1 a 3, -1 b<1, then the value range of ab is -1, 4. Answer: Solution:

1 a 3, -1 b 2, -2 b 1, -2+1 a-b 1+3, that is, -1 a-b 4, so the value range of a-b is [-1,4], and the value range of a-b can be obtained by using the xing quality of the inequality.

The title is wrong, can you take a closer look?

1 a 3, -1 b 1, -1 b 1, -1+1 a-b 1+3, i.e., -1 a-b 4, so the range of values of a-b is.

Let ac=b, ab=c=1, bc=a=2

According to the trilateral theorem of triangles, 1 b<3

It is obtained by the cosine theorem (c = a b -2abcosc).

cosc = (3+b) 4b, again 1

Solution: Let b=x know a=2 c=1 so a+c is greater than b than a-c and 3 is greater than b than 1

Using the cosine theorem cosc=(a 2+b 2-c 2) 2ab= (3+x 2) 4x=3 4x + x 4 greater than or equal to the root number 3 of half (using the fundamental inequality) when taking the equal sign x=root number 3 can be obtained.

Let f(x)=3 4x + x 4 be a simple checkpoint in the interval (1,3) and the minimum value is taken as x=root 3.

x=1 when f(x)=1 x=3 so the range of f(x) is (the root of the second half of the number 3 1).

i.e. the range of the cosc is (two-part root number 3 1).

Therefore, the value range of c is (0°, 30°).

I've done this question and I hope it helps.

There is no angle in the three corners, only two adjacent edges, and there is nothing to do with the C side, how to solve it?

Because ab+bc>c>ab-bc

Therefore, the value range of c is (1,3).

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