One angle of a triangle is 30 degrees, and the degree of the other angle is twice that of this angle. What kind of triangle is this triangle?

One angle of a triangle is 30 degrees, and the degree of the other angle is twice that of this angle. What kind of triangle is this triangle?

One angle is 30, the other is 30 * 2 = 60, and the last one is 180-30-60 = 90, so it is a right triangle
right triangle
right triangle
30*2=60°
180-30-60 = 90 degrees
One angle is 30 degrees, and the other is twice as many
So the angle is 60 degrees
So the other angle is 180 degrees - 60 degrees - 30 degrees = 90 degrees
Is a right triangle
A triangle is a () triangle if the sum of the degrees of the two angles equals the third angle
right triangle
∠A+∠B=∠C
∠A+∠B+∠C=180
Ψ∠ C = right angle
So it's a right triangle
right triangle
If f = sin (3x - π / 4) satisfies f = A and a is between 0 and 1, find the sum of all real number roots in (0,2 π)
1/3(2kπ+ arcsin a +π/4),k=0,1,2;
1/3(2kπ+ 5/4 π-arcsin a),k=0,1,2;
SUM=11/2 π
A = {x L (x + 1) (xsquare-4 = 0} B = {x L (xsquare-x-2 = 0} for intersection and union
Find a and get x ^ 2-4 = 0
x=±2
Find B and get x ^ 2-x-2 = 0
(x-2)(x+1)=0
x1=2 x2=-1
A intersection B = {x | x belongs to (- 2, - 1,2)}
A and B = {x | x = 2}
The equation of a symmetry axis of the image of the function f (x) = sin (2x + φ) is x = π 8, and φ∈ (0, π), then φ = ()
=Is it π / 4?
Because the equation of symmetry axis of F (x) = sin (2x + φ) is k π + π / 2 (k is an integer), then:
2X + φ = k π + π / 2, and x = π 8, substituting: φ = k π + π / 2-16 π (k is an integer)
And because: φ∈ (0, π), then: k = 16, that is, φ = π / 2
Let a = {(x, y) | x + 3Y = 7} and B = {(x, y) | X-Y = - 1}, then a ∩ B=______ .
If x + 3Y = 7x − y = − 1, we can get x = 1y = 2, that is, a ∩ B = {(1,2)}
Find the minimum positive period and maximum value of the function y = sin (2x + π / 6) + 2cos (2x + π / 3)
The original function is expanded by the formula of the sum of two angles,
Y = √ 3 / 2 * sin2x + 1 / 2 * cos2x + 2 (1 / 2 cos2x - √ 3 / 2 sin2x)
=3 / 2 cos2x - √ 3 / 2 sin2x congeners and union
=The common factor is extracted from √ 3 (√ 3 / 2 cos2x - 1 / 2 sin2x) and converted to auxiliary angle formula
=Application of auxiliary angle formula of √ 3cos (2x + π / 6)
So t = 2 π / 2 = π, y (max) = √ 3
The main knowledge points: sum difference formula of triangle string, auxiliary angle formula
The main thing is simplification
Let a = {x ＾ 2 / 4 + 3Y ＾ 2 / 4 = 1}, B = {y ＾ y = x ＾ 2}, then a ∩ B =?
How to solve set a? How to solve the value range of set a,
To find the intersection of set a and B is to find the intersection of two curves
Substitute y = x & # 178; into X & # 178 / 4 + 3Y & # 178 / 4 = 1
x²/4+3x²/4=1
x²=1
X = 1 or x = - 1
So a ∩ B = {(x, y) | (1,1), (- 1,1)}
3Y ^ 2 / 4 > 0, so 1 > x ^ 2 / 4 > 0,4 > x ^ 2 > 0, a = {x | 2 > x > - 2}. B = {y | Y > 0}. A intersection B = {x | 0}
Finding the period of function y = 2cos (1 / 3x - π / 4)
In the range of cosine function.
Tmin=2π/(1/3)=6π.
Cosine function
y=acos(wx+θ)(a≠0,w>0,θ∈R)
T=2π/w
T=2π/(1/3)=6π.
If the set a = {(x, y) | x-2y = 1}, B = {(x, y) | x + 3Y = 6}, then the intersection of a and B is equal to? And the union of a and B is equal to?
Two equations are connected
x-2y=1
x+3y=6
The solution is x = 3
Y=1
Then a ∩ B = (3,1)