Is there an acute angle A and B such that a + 2B = 2-sect / 3 and Tan (A / 2) * tanb = 2-radical 3 hold simultaneously Find a, b value

Is there an acute angle A and B such that a + 2B = 2-sect / 3 and Tan (A / 2) * tanb = 2-radical 3 hold simultaneously Find a, b value

Because a and B are acute angles, Tana > 0, tanb > 0, Tan (A / 2) > 0
Therefore, Tan (A / 2) tanb > 0 = 2 - √ 3
a=120-2b
tan(a/2)=tan(60-b)
tan(60-b)*tanb=2-√3
tan(60-b)=(√3-tanb)/(1+√3tanb)
(√3-tanb)/(1+√3tanb)*tanb=2-√3
(√3-tanb)tanb=(2-√3)(1+√3tanb)
√3tanb-tan^2b=2+2√3tanb-√3-3tanb
tan^2b-(3-√3)tanb+(2-√3)=0
△=4-2√3
tanb1=[3-√3+√(4-2√3)]/2
tanb2=[3-√3-√(4-2√3)]/2
Because tanb > 0
According to the test: [3 - √ 3 + √ (4-2 √ 3)] = 2
tanb1=1
[3-√3-√(4-2√3)]=4-2√3
tanb2=2-√3
Note: 3 - √ 3 ± √ (4-2 √ 3) this formula is unreasonable and special. It can not be known by manpower alone and needs to be tested by higher computer
From tanb1 = 1, B1 = 45 degrees
From tanb2 = 2 - √ 3, B2 = 15 degrees
A1 = 120-2 * 45 = 30 degrees
A2 = 120-2 * 15 = 90 degrees
Therefore, a and B have only one set of solutions, a = 30, B = 45

Is there an acute angle α, β makes α + 2 β = 2 μ g / 3 and Tan β Tan (α / 2) = 2-radical 3? Request alpha, beta

Therefore, Tan (α / 2) + Tan β = 3 - √ 3. Therefore, Tan (α / 2) and Tan β are the roots of the quadratic equation x ^ 2 - (3 - √ 3) x + (2 - √ 3) = 0. The two roots of this equation are positive numbers 1 and 2 √ 3
When Tan (α / 2) = 1, α = π / 2. So tan (α / 2) = 2 - √ 3, Tan α = 1 / √ 3, so α = π / 6, β = π / 4

Is there an acute angle α and β such that ① α + 2 β = 2 π / 3; ② Tan α / 2 * Tan β = 2 - √ 3? If there is, calculate the value of α and β; if not, explain the reason

T α n (α / 2 + β) = [t α n (α / 2) + T α n β] / [1-T α n (α / 2) t α n β] means that t α n (π / 3) = [t α n (α / 2) + T α n β] / [1 - (2 - √ 3)] = √ 3T α n (α / 2) + T α n β (β) n β = √ 3 (√ 3-1) t α n (α / 2) + T α n β = 3 - √ 3, we can know (t α n α / 2) and t α n β is the equation x ^ 2 - (3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-(x + 2 - √ 3 =

It is known that α β is an acute angle and the sine value of α is 4 √ 3 / 7. The cosine value of α + β is equal to - 11 / 14. Find the sine value of β

α. β is an acute angle, 0 < α + β < π,
——》Cos α and sin (α + β) were higher than 0,
——》cosα=v(1-sin^2α)=1/7,sin(α+β)=v[1-cos^2(α+β)]=5v3/14,
——》sinβ=sin[(α+β)-α]
=sin(α+β)cosα-cos(α+β)sinα
=(5v3/14)*(1/7)-(-11/14)*(4v3/7)
=v3/2.

If a is an acute angle and cosine a equals four fifths, what is the value of sine 90 degrees minus a

Because the cosine is four fifths, so the hypotenuse is 5 and the other side is 4. He tells you that the sine is 90 degrees, so the third side of Pythagorean theorem is 3

The sine value of an acute angle is equal to the sine value of the complement angle of the acute angle Is it true

yes
sina=sin(180-a)

It is proved that the cosine of the acute angle between the two right sides of isosceles right triangle is 4 / 5 The proof didn't work out for a long time

1. In the isosceles right triangle ABC, B is the vertex of the right angle, e and F are the midpoint of the BC side and ab side. Connect AE and CF, and let the intersection point be o; 2. If AB = BC = 2, then AF = BF = be = CE = 1, then the length of AE and CF can be calculated. 3

Calculate sin30 ° - cos45 ° / 1 + tan60 ° = (keep root)

sin30°-cos45°÷(1+tan60°)
=1/2-√2/2÷1+√3
=1/2-√2/2+√3

If ∠ x is an acute angle in a right triangle and Tan x is equal to 3, what is cos x equal to

Let two right angles be a, B. the hypotenuse is C. then TaNx = A / b = 3, that is, a = 3B ~ (1)
According to Pythagorean theorem, a ^ 2 + B ^ 2 = C ^ 2 ~ (2)
From (1) (2), C ^ 2 = 10B ^ 2 ~ (3)
Because x is an acute angle, cosx is a positive number. From (3) (4), cosx = √ 10 / 10

It is known that α and β are acute angles, cos α 1 7，cos（α+β）=-11 14, then cos β = 0___ .

α. β is acute angle,
∴sinα=
1-1
49=4
Three
7，sin（α+β）=
1- (11
14)  2=5
Three
Fourteen
∴cosβ=cos[（α+β）-α]=cos（α+β）cosα+sin（α+β）sinα=1
2．
So the answer is 1
Two