If α and β are acute angles, and sin α = 5 / 5 under the root sign, Tan β = 1 / 3, then α + β =?

5 / 5 cos α = 2 √ 5 / 5 cos α = 2 √ 5 / 5 Tan β = 1 / 3 sin β = (10 / 10 cos β = 3 √ 10 / 10) 10 / 10sin (α + β) = sin α cos β + cos α α sin β = 3 √ 5 / 5 * 3 √ 10 / 10 + 2 √ 5 / 5 * √ 5 / 5 * √ 10 / 10 = 10 / 10 = 3 √ 50 / 50 + 2 √ 50 / 50 = 5 √ 50 / 50 = 25 √ 50 / 50 = 25 √ 2 / 2 α + β = 45 ° or 135 ° because of Tan β = 1 / 1 / 1 / 135 ° because of Tan β = 1 / 1 / 1 / 1 / 1 / 1 / 1 / 135 ° because Tan β = 1 / 1 / 1 / 1 / 3, β

Find the size of the acute angle α, Tan 2 α (root 3 + 1) Tan α + root 3 = 0

Let Tan α = XX ^ 2 - (√ 3 + 1) x + √ 3 = 0. (x - √ 3) (x-1) = 0, X - √ 3 = 0, X1 = √ 3; X-1 = 0, X2 = 1, νx1 = Tan α = √ 3, α 1 = 60 ° x2 = Tan α = 1. α 2 = 45 ° α = 60 ° or α = 45 °

Is there an acute angle A and β such that 1. A + 2 β = 2 * 180 / 3; 2. Tana / 2 * Tan β = 2-radical 3 holds simultaneously

Is there an acute angle A and β such that 1. α + 2 β = 2 * 180 / 3; 2. Tana / 2 * Tan β = 2-radical 3 holds at the same time? If so, the known condition 1. α + 2 β = 2 * 180 / 3 = 120 ° = > α / 2 + β = 180 / 3 = 60 ° Tan (α / 2 + β) = [Tan (α / 2) + Tan (β)] / [1-tan (α / 2

If ∠ α is an acute angle and 3 (Tan α) ^ 2-4 root sign 3 (Tan α) + 3 = 0, then the degree of α?

3 (Tan α) ^ 2-4 root sign 3 (Tan α) + 3 = 0
(3tan α - radical 3) (Tan α - radical 3) = 0
Tan α = root 3 / 3, or tan α = root 3
The acute angle is α
α = 30 ° or α = 60 °

It is known that α is an acute angle and Tan (90 ° - α) = radical 3

From Tan (90 ° - α) = radical 3
90 ° - α = (PI) / 3 + K (PI) (PI is PI)
When α is an acute angle, α = pi / 6 (or α = 30 °)

2sin50°+cos10°(1+ 3tan10°) 1+cos10°=______ .

The original formula = 2sin50 ° + cos10 ° · cos10 °+
3sin10°
cos10°
2cos5°=2sin50°+2sin40°
2cos5°=2
2sin(50°+45°)
2cos5°=2
2cos5°
2cos5°=2．
So the answer is: 2

[2sin50 + sin10 (1 + radical 3 * tan10)] * Radix (2sin80 * sin80)

sqrt(3)=tan60
1+sqrt(3)*tan10=1+tan60tan10=(sin60sin10+cos60cos10)/cos60cos10=cos50/cos60cos10=2cos50/cos10
sqrt(2sin80*sin80)=sqrt(2)sin80=sqrt(2)cos10
[2sin50+sin10(1+sqrt3*tan10)]*sqrt(2sin80*sin80)=[2sin50+2cos50sin10/cos10]sqrt(2)cos10=2sqrt(2)*(sin50cos10+cos50sin10)=2sqrt(2)*sin60=sqrt(6)

[2sin50 degree + cos10 Degree * (1 + root 3 * Tan 10 degree)] / (root 2 * cos 5 degree)

The original formula = (2sin50 ℃ + cos10 ℃ + √ 3sin10 ℃) / √ 2cos5: = [2sin50 ° + 2 (1 / 2 * cos10 ° + √ 3 / 2sin10 ℃) / √ 2cos5: = (2sin50 ° + 2cos50 ℃) / √ 2cos5: = 2 √ 2 (√ 2 / 2sin50 ° + √ 2 / 2CO

2 sin 50 ° + (1 + radical 3 * tan10 °) cos10 ° / Radix 2 / cos5 ° were calculated

Find the square value of 2 times sin80 degree under the root sign [2sin50 degree + sin10 degree (1 + root sign 3 * tan10 degree)] *

Radical 6 formula = [2sin50 + 2sin10 / cos10 (1 / 2 * cos10 + root 3 / 2 * sin10)] * Radix 2sin80 = [2sin50 + 2sin10 cos10 (sin30cos10 + cos30 sin10)] * Radix 2sin80 = (2sin50 + 2sin10cos10sin40) * 2sin80 (sin 40 = cos 50) = 2 / cos10 (sin50 cos1

If α and β are acute angles, and sin α = 5 / 5 under the root sign, Tan β = 1 / 3, then α + β =?