In the fourth quadrant (a), then (a) |

In the fourth quadrant (a), then (a) |

A is in the fourth quadrant
It can be seen that 270 + n * 360

Cosa = - 1 / 3, a is the second quadrant angle, and sin (a + b) = 1, find the value of COS (2a + b)

Cosa = - 1 / 3, a is the second quadrant angle, Sina = 2 √ 2 / 3,
sin(a+B)=1,cos(a+B)=0
cos(2a+B)=cos[(a+B)+a]
=cos(a+B)cosa-sin(a+B)sina
=-2√2/3
or
sin(a+B)=1,a+B=2kπ+π/2
cos(2a+B)=cos[(a+B)+a]
=cos(2kπ+π/2+a)
=cos(π/2+a)
=-sina
=-2√2/3

If a is an acute angle, what quadrant angle is 2A?

Because a is an acute angle
So 0 ° < a < 90 °
0° < 2a < 180°
So in the first or second quadrant

It is known that a is an acute angle and satisfies √ 3tan ^ 2a-4tana + √ 3 = 0 Find degree a

√3tan²a-4tana+√3=0
(√3tana-1)(tana-√3)=0
√ 3tana-1 = 0 or Tana - √ 3 = 0
Tana = √ 3 / 3 or Tana = √ 3
A = 30 ° or a = 60 '

In an acute triangle, edges a, B and C correspond to angles a, B and C respectively. Given a = 1 and B = 2A, find the value of B / cosa and the range of B

a/sinA=b/sinB
sinB=2sinAcosA
So a / Sina = B / (2sinacosa)
b=2*cosA*a=2cosA
So B / cosa = 2
Because B = 2A, a belongs to (0,60 °)
Cosa belongs to [0.5,1)
B belongs to [1,2]

(1) Given that a is an acute angle, if 2Sin ^ 2A = 1, find the value of A (2) Given that a is an acute angle, if the root of the equation 3x ^ 2-6x + 9tana = O has two real roots equal to, find the value of A

α is an acute angle, sin α > 0
sin^2a=1/2
Sin α = radical 2 / 2 α = 45 degrees
The equation has two equal real roots
Delta = 0, that is (- 6) ^ 2-4 * 3 * 9tan α = 0
α is an acute angle, Tan α > 0
tanα=1/3
α=rac tan1/3

It is known that both a and B are acute angles, then the range of angle 2a is______ The range of A-B angle is______ ?

∵0° ∵ 0 degree < a < 90 degree, 0 degree < B < 90 degree and - 90 degree < - B < 0 degree, so - 90 degree < A-B < 90 degree

In acute triangle, 2A = root 3b, find angle A

There is no solution to the problem and the conditions are not enough

It is known that a is the first quadrant angle, 2a is the quadrant angle, and a / 2 is the quadrant angle

0＜a＜90,
0 < 2A < 180, so 2a is in the first or second quadrant
0 < A / 2 < 45, so a / 2 is in the first quadrant

The symmetric point of point P (a + 1, 2a-1) about the x-axis is known to be in the first quadrant, and the value range of a is calculated

According to the meaning of the question, point P is in the fourth quadrant,
Qi
a+1＞0
2a−1＜0 ，
The solution is: - 1 < a < 1
2，
That is, the value range of a is - 1 < a < 1
2．