Find B / a if C ^ 2 = B ^ 2 + root 3A ^ 2, find B

Find B / a if C ^ 2 = B ^ 2 + root 3A ^ 2, find B

According to the sine theorem a = 2rsina, B = 2rsinb, where R is the diameter of circumscribed circle, we can get 2rsinasinb + 2rsinb (COSA) ^ 2 = √ 2 * 2rsina [(Sina) ^ 2 + (COSA) ^ 2] SINB = √ 2sinasinb / Sina = √ 2. According to the cosine theorem B ^ 2 = C ^ 2 + A ^ 2-2accosb, the original formula is transformed into B ^ 2 =

If a, B, C are the three sides of the triangle ABC, the absolute value of the radical (a-b-c) 2 + B + a-c is simplified

The sum of the two sides of the triangle is greater than the third
So a-b-c0
So the original formula = | a-b-c | + | B + a-c|
=-a+b+c+b+a-c
=2b

It is known that the three sides of △ ABC are a, B and C respectively and satisfy the following conditions A − 1 + B 2 − 4B + 4 = 0, find the value range of the third side C

A kind of
a−1+b2-4b+4=
a−1+（b-2）2=0，
A-1 = 0, B-2 = 0, that is, a = 1, B = 2,
Then the range of the third side C is 2-1 < C < 2 + 1, i.e. 1 < C < 3

It is known that the three sides of △ ABC are a, B and C respectively and satisfy the following conditions A − 1 + B 2 − 4B + 4 = 0, find the value range of the third side C

A kind of
a−1+b2-4b+4=
a−1+（b-2）2=0，
A-1 = 0, B-2 = 0, that is, a = 1, B = 2,
Then the range of the third side C is 2-1 < C < 2 + 1, i.e. 1 < C < 3

The three sides of the triangle ABC are a, B and C respectively, and a and B satisfy the root sign A-1, the square of B minus 4B plus 4 equals 0, so we can find the value range of C

√(a - 1) + (b² - 4b + 4) = 0
√(a - 1) + (b - 2)² = 0
Because the root is greater than or equal to 0, and the square is greater than or equal to 0
So a - 1 = 0 and B - 2 = 0
So a = 1, B = 2
Because the three sides of the triangle ABC are a, B, C
Therefore, C + C > a > b > C + C + b > a > b > C + b > a > b > a > b > a > b > a > b > a > b > a > b > a > b > a > b > a > b > a > b > a > b > a >
The solution is: 1 < C < 3

It is known that the three sides of △ ABC are a, B and C respectively and satisfy the following conditions A − 1 + B 2 − 4B + 4 = 0, find the value range of the third side C

A kind of
a−1+b2-4b+4=
a−1+（b-2）2=0，
A-1 = 0, B-2 = 0, that is, a = 1, B = 2,
Then the range of the third side C is 2-1 < C < 2 + 1, i.e. 1 < C < 3

Given the triangle ABC's three side length a B C, where AC satisfies A-1 + the square of C under root sign - 4C + 4 = 0, find the value range of B We know the three side length a B C of the triangle ABC, where a and C satisfy the square of A-1 + C under the root sign - 4C + 4 = 0, and find the value range of B

If you can't type the root sign, use words to explain it
(A-1) + C ^ 2-4c + 4) = 0
Under root sign (A-1) + root sign ((C-2) ^ 2) = 0
Under radical sign (A-1) + | C-2 | = 0
a-1=0 ,c-2=0
So a = 1, C = 2
Because B satisfies | a-c|

If a, B, C are the three sides of △ ABC, then the result of simplifying | a-b-c | + | b-c-a | + | C-A-B |, is () A. -a-b-c B. a+b+c C. a+b-c D. a-b+c

∵ a, B, C are the three sides of ∵ ABC,
∴a＜b+c，b＜c+a，c＜a+b，
∴a-b-c＜0，b-c-a＜0，c-a-b＜0，
∴|a-b-c|+|b-c-a|+|c-a-b|
=b+c-a+c+a-b+a+b-c
=a+b+c．
Therefore, B

It is known that the three sides of the triangle ABC are ABC respectively and ABC satisfies the absolute value of a 2 - Ba + 9 + radical B-4 + C-5. Try to judge the shape of the triangle ABC

If a? - 6A + 9 + √ (B-4) + | C-5 | = 0, i.e. (A-3) 2 + √ (B-4) + | C-5 | = 0, then A-3 = 0, B-4 = 0, C-5 = 0, ? a = 3, B = 4, C = 5, ∵ 3 ∵ 3 ∧ 4 ∵ 4 ∵ 3 ∵ 4 ∵ 3 ∵ 3 ∵ 3 ∵ 3 ? 3 ∵ 4 ?

In the triangle ABC, ABC is the opposite side length of angle ABC, s is the area of triangle ABC, and 4sinbsin 2 (4 / π + 2 / b) + cos2b = 1 + radical 3 (1) Find the degree of angle B (2) If a = 4, s = 5, root sign 3, find the value of B

1. Ask, is it 4sinbsin? 2 (π / 4 + B / 2) + cos2b = 1 + radical 3?
2 SINB [1-cos (π / 2 + b)] + cos2b = 1 + radical 3
Continue to reduce SINB = 1 / 2 radical 3
So B = π / 3 or 2 π / 3
2. S = 1 / 2acsinb is brought into the data to obtain C = 5
According to cosine theorem, B ^ 2 = a ^ 2 + C ^ 2-2accosb
And CoSb = 1 / 2 or - 1 / 2, so B = root 21 or root 61
Hope my answer can give you some help, thank you!