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

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, the ratio of degree of the outer angle of ∠ a ∠ B ∠ C is 2:3:4. Find the degree of ∠ a
The sum of the external angles of the triangle is 360 ° and the complementary angle of ∠ a can be calculated as 80 °
The sum of ∠ A and its complementary angle is 180 degrees, so ∠ A is 100 degrees
If the sum of internal angles of a triangle is 180 degrees, then the sum of external angles is 180 (360-180), so let x, get 2x + 3x + 4x = 180, x = 20, and angle a is 2x, so it is 40 degrees
If 2x of a is equal to 2 / 3, then (3x of a plus - 3x of a) / (x of a plus - X of a) is equal to what?
(3x of a plus - 3x of a) / (x of a plus - X of a) = (a ^ x + A ^ - x) (a ^ 2x-a ^ x * a ^ (- x) + A ^ (- 2x)) / (a ^ x + A ^ - x) = a ^ 2x-a ^ x * a ^ (- x) + A ^ (- 2x) = a ^ 2x-1 + A ^ (- 2x) because the 2x of a is equal to 2 / 3a and the - 2x of a is equal to 3 / 2, the original formula = 2 / 3-1 + 3 / 2 = 7 / 6
It is proved that 2Sin (π / 4 + x) sin (π / 4-x) = cos2x
2sin(π/4-x)sin(π/4+x)
=2(sinπ/4cosx-cosπ/4sinx)(sinπ/4cosx+cosπ/4sinx)
=2[(sinπ/4cosx)^2-(cosπ/4sinx)^2]
=2[(cosx)^2/2-(sinx)^2/2]
=(cosx)^2-(sinx)^2
=cos2x
What process is (3x + 2) (2x + 1) (x-1) equal to
　(3x+2)(2x+1)(x-1)
=(6x²+3x+4x+2)(x-1)
=(6x²+7x+2)(x-1)
=6x³+7x²+2x-6x²-7x-2
=6x³+x²-5x-2
If cos2x / sin (x - π / 4) = - √ 2 / 2, then the value of √ 2Sin (x + π / 4) is
Let x + π / 4 = a
cos2x/sin(x-π/4)=-√2/2,
That is cos [2A - π / 2] / sin (a - π / 2) = - √ 2 / 2,
We get sin [2A] / cosa = √ 2 / 2,
2sina==√2/2,
sina==√2/4
√2sin(x+π/4)=√2sina=1/2
Two thirds root two
Let the elements in the set G be all numbers of the form a + B √ 2 (a ∈ Z, B ∈ z), and prove 1. When x ∈ Z, X ∈ G
Is the expression of X correct
The image of function y = cos (3x + π / 3) can be obtained from the image of y = SiNx
Y = SiNx image → shift 5 π / 6 unit length to the left → y = sin [x + 5 π / 6] image → y = sin [(x + π / 3) + π / 2] image → y = cos (x + π / 3) image → ordinate unchanged, abscissa reduced to original 1 / 3 → y = cos (3x + π / 3) image or y = SiNx image → ordinate unchanged, abscissa reduced
First shift π / 2 to the left, then decrease to 1 / 3 times in the cycle, and then shift π / 9 to the left
1. The image is shifted to the right by π / 2 units, y = cosx;
2. The image is compressed horizontally to 1 / 3 of the original image, y = cos3x;
3. The image wants to shift 1 / 9 unit to the left, y = cos [3 (x + π / 9)] = cos (3x + π / 3)
One
y=sinx
The period becomes 1 / 3 of the original, that is, the abscissa of all points becomes 1 / 3 of the original,
Become
y=sin3x
Two
Because y = cos (3x + π / 3)
=sin(3x+π/3+π/2)
=sin(3x+5π/6)
=sin3(x+5π/18)
therefore
As long as y = sin3x is moved 5 π / 18 units to the left, the image of y = cos (3x + π / 3) can be obtained.
remainsb is right！
y=cos(3x+π/3)=sin[π/2-(3x+π/3）]=sin(π/6-3x)=-sin（3x-π/6）。 The image of y = SiNx is translated by π / 6 units along the x-axis, then the abscissa is shortened to 1 / 3 of the original, and finally the image is folded about the x-axis (the x-axis is regarded as the axis of symmetry).
On the transformation law of trigonometric function image is limited to space, you can refer to the chapter of trigonometric function in high school textbook. ... unfold
y=cos(3x+π/3)=sin[π/2-(3x+π/3）]=sin(π/6-3x)=-sin（3x-π/6）。 The image of y = SiNx is translated by π / 6 units along the x-axis, then the abscissa is shortened to 1 / 3 of the original, and finally the image is folded about the x-axis (the x-axis is regarded as the axis of symmetry).
On the transformation law of trigonometric function image is limited to space, you can refer to the chapter of trigonometric function in high school textbook. Put it away
y = cos(3x+π/3)
= cos[3(x+π/9)]
= sin[3(x+π/9)+π/2]
= sin[3(x+π/9+π/6)]
= sin[3(x+5π/18)]
The drawing is as follows:
(1) Make the image of y = SiNx;
(2) The value of X in the image of y = SiNx is reduced to 1 / 3 of the original value. The graph expansion of y = sin3x is obtained
y = cos(3x+π/3)
= cos[3(x+π/9)]
= sin[3(x+π/9)+π/2]
= sin[3(x+π/9+π/6)]
= sin[3(x+5π/18)]
The drawing is as follows:
(1) Make the image of y = SiNx;
(2) The value of X in the image of y = SiNx is reduced to 1 / 3 of the original value. We get the image of y = sin3x.
(3) The whole image of y = sin3x is shifted 5 π / 18 to the left. Put it away
Given x + y + 13-quadruple root x-sextuple root y = 0, find the value of real number XY
X + y + 13-quadruple radical x-sextuple radical y = 0
It is equivalent to (radical X-2) ^ 2 + (radical Y-3) ^ 2 = 0, so x = 4, y = 9, xy = 36
How to solve the definite integral ∫ (2x + 3x-4) DX
∫(2^x +3^x-4)dx
=∫2^xdx +∫3^xdx-4∫dx
=2^x/ln2 +3^x/ln3-4x+C
Is something missing? Is x ^ 2 followed by 2?
∫(2x +3x-4)dx=2/3x^3+3/2x^2-4x+c
This problem should not be difficult. Let's see if the formula can work out.