In the triangle ABC, if 2 (angle a + angle c) = 3 angles B, then the degree of the outer angle of angle B

In the triangle ABC, if 2 (angle a + angle c) = 3 angles B, then the degree of the outer angle of angle B

In triangle ABC, 2 (angle a + angle c) = 3 angles B
∴∠A+∠C=1.5∠B
∵∠A+∠C+∠B=180°
∴2.5∠B=180°
∠B=72°
The external angle degree of angle B = 180 ° - 72 ° = 108 °
That is, 2 (180-b) = 3B
360-2b=3b
b=72
Then 180 - no 08
So the outside angle is 108 degrees
∵ angle a + angle c = 180 ° - angle B
Substituting into the original formula, 2 (180 ° - b) = 3 ∠ B
∴∠B=72°
The outer angle of ℅ B is 108 degrees
In △ ABC, if the ratio of the degrees of the outer angles of ∠ a, ∠ B, ∠ C is 4:3:2, find the degree of ∠ a
Suppose that the external angles of ∠ a, B and C are ∠ 1 = 4x degree, ∠ 2 = 3x degree and ∠ 3 = 2x degree respectively. (1 point) because ∠ 1, ∠ 2 and ∠ 3 are the three external angles of △ ABC, so 4x + 3x + 2x = 360, the solution is x = 40. (2 points) so ∠ 1 = 160 °, 2 = 120 °, 3 = 80 ° (1 point) because ∠ a + ∠ 1 = 180 ° and (1 point) so ∠ a = 20 ° (1 point)
In the triangle ABC, ∠ A: ∠ B: ∠ C = 1:2:3, then the degree ratio of the outer angle of the triangle is____
Teach me, just answer
The three internal angles are 30, 60 and 90
The three external angles are 150, 120 and 90
The ratio was 150:120:90 = 5:4:3
5：4: 3
11：10：9
5:4:3
5:4:3
a: B: C = 1:2:3, so a = 30 ° B = 60 ° C = 90 ° (because a: B: C = 1:2:3, the sum of the inner angles of the triangle is 180 °, a accounts for 6 / 1, B is 6 / 2, which is 3 / 1, C6 / 3 is 2 / 1, which is multiplied by 180 ° to get the degree of ABC respectively)
So the outer angle of a is 150 ° B is 120 ° C and the outer angle of a is 90 °
So the angle ratio of the triangle is 5:4:3
The minimum positive period of the function y = sin (2 π / 3x + π / 4) needs detailed process
Solution: because the function period T = 2 pies / 2 pies / 3 = 3, and 2 pies > 3, the minimum positive period T = 3
Given the set a = {x ax + 2A + 6 ＜ 0}, B = {x x ＜ 0}. If a ∩ B = & Oslash; (empty set), find the value range of real number a
a> 0, there is a: x = 0
-3=
The minimum positive period of the function y = sin 3x is
2∏/3
Given the set a = {x | x2-3x + 2 = 0}, B = {x | x2-2x + A-1 = 0}, B belongs to a, the range of finding x?
It is the range of ball a! From the solution of a: x ^ 2-3x + 2 = 0, x = 1, x = 2, a = {1,2} B belongs to a = {1,2} B, which can be substituted by {0}, {1}, {2} to get a = {1,2}
Using group decomposition method 5x ^ 2 + 6y-15x-2xy x ^ 3 + 9 + 3x ^ 2 + 3x 2x ^ 2 + xy-y ^ 2-4x + 5y-6
5x^2+6y-15x-2xy
=（5x²-15x）+（6y-2xy）
=5x（x-3）+2y（3-x）
=（5x-2y）（x-3）
x^3+9+3x^2+3x
=（x³+3x²）+（3x+9）
=x²（x+3） +3（x+3）
=（x²+3）（x+3）
2x^2+xy-y^2-4x+5y-6
This problem can be looked at separately
First, it is divided into 2x & # 178; - 4x-6 - Y & # 178; + 5y-6
=（2x-6）（x+1） =-（y²-5y+6）
=（y-2）（y-3）
There is clearly no way to merge
So, you have the wrong number
Find the integer solution that satisfies 2-3x greater than or equal to 2x-8 and 1 / 2-x less than 2-x / 3 + 1 at the same time
2-3x≥2x-8
1/2-x
23456789, all of them
Simplification: cos4x * cos2x cos ^ 2 * 3x
Original formula = 1 / 2 * (cos6x + cos2x) - 1 / 2 (1 + cos6x)
=(cos2x-1)/2
=(1-2sin²x-1)/2
=-sin²x