In triangle ABC, angle a-angle C = 35 ° and angle B-angle a = 5 °, then angle B =?

In triangle ABC, angle a-angle C = 35 ° and angle B-angle a = 5 °, then angle B =?

The sum of the internal angles of a triangle is equal to 180 ° so three equations can be established
A+B+C=180 （1)
A-C=35 ->A=35+C （2)
B-A=5 （3)
(2) The formula of + (3) is obtained
B-C=40 ->B=40+C (4)
Substituting (2) (4) into (1) has:
35+C+40+C+C=180 ->C=35
So
A＝70
B＝75
C＝35
So angle B is 75 degrees

In the triangle ABC, ∠ C = 90 °, a = 6, C = 10, then the area of triangle ABC is

∵∠C＝90°,
A & # 178; + B & # 178; = C & # 178; (Pythagorean theorem)
Substituting a = 6 and C = 10, we get the following result:
36＋b²＝100
The solution is: B = 8
The area of RT △ ABC is 6 × 8 △ 2 = 24

Proof: the product of four consecutive integers plus 1 is the square of an integer

Let the four consecutive integers be n-1, N, N + 1, N + 2, then (n-1) n (n + 1) (n + 2) + 1, = [(n-1) (n + 2)] [n (n + 1)] + 1 = (N2 + n-2) (N2 + n) + 1 = (N2 + n) 2-2 (N2 + n) + 1 = (N2 + n-1) 2. Therefore, the product of four consecutive integers plus 1 is the square of an integer

The product of multiplication of two odd numbers must be______ ．

According to the above analysis: odd × odd = odd, such as: 3 × 5 = 15, 3 × 7 = 21; answer: if two odd numbers multiply, the product must be odd. So the answer is: odd

Given that ABC is a positive integer, prove a ^ 3 + B ^ 3 + C ^ 3 + 1 / ABC > = 2 radical 3

Original formula = (3a & # 179; + 3B & # 179; + 3C & # 179; + 3 / ABC) / 3
= (3a³+3b³+3c³+1/abc+1/abc+1/abc)/3
≥ 6*(3a³*3b³*3c³*1/abc*1/abc*1/abc)^(1/6) / 3
= 6*√3 / 3
= 2√3

ABC = 1, prove that 1 / A + 1 / B + 1 / C is greater than or equal to radical a + radical B + radical C

Certification:
1/a+1/b+1/c=(ab+bc+ac)/abc
=ab+bc+ac
=(1/2)[(ab+bc)+(ab+ac)+(ac+bc)]
≥(1/2)[2(ab*bc)^(1/2)+2(ab+ac)^(1/2)+2(ac+bc)^(1/2)]
=(abc*b)^(1/2)+(abc*a)^(1/2)+(abc*c)^(1/2)
=b^(1/2)+a^(1/2)+c^(1/2)
It has been proved

It is known that a, B, C ∈ R +, a, B, C are not equal to each other and ABC = 1. Proof: a + B + C < 1A + 1b + 1C

It is proved that: ∵ a, B, C ∈ R + are not equal to each other, and ABC = 1 ∵ a + B + C = 1BC + 1Ac + 1ab < 1b + 1C2 + 1A + 1C2 + 1A + 1b2 = 1A + 1b + 1C

Given that a, B, C are mutually unequal positive numbers, and ABC = 1, prove: radical a + radical B + radical C

On the right, change 1 to ABC
So the right side = BC + AC + AB = 1 / 2 * (2BC + 2Ac + 2Ab) = 1 / 2 * [(AB + AC) + (Ba + BC) + (Ca + CB)] = 1 / 2 * [a (B + C) + B (a + C) + C (a + b)] > = 1 / 2 * [a * 2 * radical (BC) + b * 2 * radical (AC) + C * radical (AB)] = a * radical (BC) + b * radical (AC) + C * radical (AB) = a * radical (1 / a) + b * radical (1 / b) + C * radical (1 / C) = a + radical (B + radical C)
Because ABC is not equal, a + C > = 2 * radical AC, B + C > = 2 * radical BC, B + a > = 2 * radical Ba are obtained at the same time, so the original formula is proved
The key is the substitution of 1, which is often used in high school inequality proof You need to be proficient There are also important inequalities to keep in mind

It is known that the lengths of three sides of △ ABC are a, B, C respectively, and (a-2b + 1) & sup2; + B-3 = 0, C is a positive integer

1.
y=x3－3,
The arithmetic square root of Y is 4,
y=16
x³-3=16
x³=19
x=(19)^(1/3)
two
a2－6a+9+√（b-4 )|c-5|=0
(a-3)²+√（b-4 )|c-5|=0
(a-3)²>=0
√（b-4 )>=0
|c-5|>=0
If the sum of three numbers greater than or equal to zero is 0, then (A-3) & sup2; = 0
√（b-4 )=0
|c-5|=0
a=3,b=4,c=5
3²+4²=5²
According to the inverse theorem of Pythagorean theorem, △ ABC is a right triangle

It is known that the lengths of three sides a, B and C of △ ABC are all positive integers, and a ≤ B ≤ C. if B is a constant, then the number of △ ABC satisfying the requirement is ()
A. b2B. 23b2+13C. 12b2+12bD. 23b2+13b

When a = 1, C can only take B, when a = 2, C can take B, B + 1; when a = 3, C can take B, B + 1, B + 2 When a = B, C can be B, B + 1, B + 2 So the number of triangles is 1 + 2 + 3