# In triangle ABC, C = 60 degrees, a + B = 2 (root 3 + 1), C = 2, root 2, then a is equal to? In triangle ABC, C = 60 degrees, a + B = 2 (root 3 + 1), C = 2, root 2, then what is a equal to? 45 degrees or 75 degrees

## In triangle ABC, C = 60 degrees, a + B = 2 (root 3 + 1), C = 2, root 2, then a is equal to? In triangle ABC, C = 60 degrees, a + B = 2 (root 3 + 1), C = 2, root 2, then what is a equal to? 45 degrees or 75 degrees

According to the cosine theorem:
C ^ 2 = a ^ 2 + B ^ 2-2abcosc = a ^ 2 + B ^ 2-AB = 8, a ^ 2 + B ^ 2 = 8 + ab
a+b=2(√3+1)
A ^ 2 + B ^ 2 + 2Ab = 8 (2 + √ 3), that is, 8 + AB + 2Ab = 8 (2 + √ 3),
So AB = 8 (√ 3 + 1) / 3
The solution is a = 4 √ 3 / 3, B = (2 √ 3 + 6) / 3
Or B = 4 √ 3 / 3, a = (2 √ 3 + 6) / 3
When a = 4 √ 3 / 3
a/sinA=c/sinC
Get a = 45 degrees, B = 75 degrees
When a = (2 √ 3 + 6) / 3
A = 75 degrees, B = 45 degrees

### In triangle ABC, a is equal to root 3, B = 1, C = 2, then

Pythagorean theorem shows that it is a right triangle

### (13 / 2 minus root sign 42) open root sign simplification 18500 look at three root signs

13/2 - √42 = (13- 2√42 )/2 = [ 7 - 2√(7*6) +6 ] /2 = (√7-√6) ² / √2 ² √(13/2 - √42) = (√7-√6) /√2 = √14 /2 - √3

solution
√8 × √2-5
=2√2 × √2-5
=2 × 2-5
=4-5
=-1

### If a > 0 is known, verify the root sign (a ^ 2 + 1 / A ^ 2) - root sign 2 > = (a + 1 / a) - 2

Let x = a + 1 / A
Then x ²= a ²+ 1/a ²+ two
Therefore, it is proved that √ (x ²- 2)-√2>=x-2
I.e. proof √ (x) ²- 2)+2>=x+√2
That is to prove [√ (x ²- 2)+2] ²>= (x+√2) ²
Proof (x) ²- 2)+4√(x ²- 2)+4>=x ²+ 2√2x+2
I.e. proof 4 √ (x) ²- 2)>=2√2x
I.e. proof 2 √ (x) ²- 2)>=√2x
That is to prove [2 √ (x ²- 2)] ²>= (√2x) ²
Proof 4x ²- 8>=2x ²
That is, prove X ²>= four
Because a > 0
So x = a + 1 / a > = 2 √ (a * 1 / a) = 2
So x ²>= 4 Establishment
Push back
Yes √ (a) ²+ 1/a ²)- √2>=(a+1/a)-2

### The size of a + B and 2 radical AB (a ≥ 0b ≥ 0) and explain the reasons Try to use only open cube, open square, cube and square

Because (√ a - √ b) ^ 2 ≥ 0,
So A-2 √ AB + B ≥ 0,
So a + B ≥ 2 √ ab