In the triangle ABC, angle c + angle a = 2, angle B, angle c - angle a = 80 degrees, calculate the angles a, B, C for several degrees and tell how to calculate them

In the triangle ABC, angle c + angle a = 2, angle B, angle c - angle a = 80 degrees, calculate the angles a, B, C for several degrees and tell how to calculate them

Because angle a + angle B + angle c = 180, and angle c + angle a = 2 angle B, so 3 angle B = 180, angle B = 60 degrees; because angle c-angle a = 80 degrees, angle c + angle a = 120 degrees, angle a = 20 degrees, angle c = 100 degrees

In triangle ABC, angle a + angle B = 2 angle c, angle B - angle a = 80 degrees, then angle B=

Angle a = 100 ° angle a = 20 ° angle c = 60 °
The first one is corner B. It's wrong. I'm sorry
Let the angle a = x, the angle B = y, so the angle c = 180 ° - (x + y)
So there is a system of quadratic equations of two variables
x+y=[180-（x+y）]x2
y-x=80
The solution is x = 20. Y = 80

In the triangle ABC, the angle a is equal to 1 / 2 and the angle B is equal to 1 / 3 angle C

Angle a equals 1 / 2 angle B equals 1 / 3 angle c b = 2A C = 3A a + B + C = 6A = 180 a = 30 B = 60 C = 90
Hope to adopt

In the known triangle ABC, ∠ C = 80 ° and ∠ a - ∠ B = 40 ° is known. Find the degree of ∠ B

Let ∠ B be x degree, then ∠ a be x + 40 degree
x+x+40+80=180
       2x=180-120
    2x×2=60÷2
        x=30
Answer: the degree of ∠ B is 30 degrees

In the known triangle ABC, ∠ C = 80 ° and ∠ a - ∠ B = 40 ° is known. Find the degree of ∠ B

Let ∠ B be x degree, then ∠ a be x + 40 degree
x+x+40+80=180
       2x=180-120
    2x×2=60÷2
        x=30
Answer: the degree of ∠ B is 30 degrees

In the known triangle ABC, ∠ C = 80 ° and ∠ a - ∠ B = 40 ° is known. Find the degree of ∠ B

Let ∠ B be x degree, then ∠ A is x + 40 ° x + X + 40 + 80 = 180 2x = 180-120 2x × 2 = 60 ÷ 2 x = 30 answer: the degree of ∠ B is 30 degrees

Given the triangle ABC and triangle def, angle a = 60 °, angle B = 40 °, angle d = 80 °, then the degree of angle E

The similarity of triangles corresponds to the equality of angles. In triangle ABC, we can find C = 80 = D, from which we can get e = b = 40 or E = a = 60

In triangle ABC, given the angle a + angle c = 2 angle B, calculate the degree of angle B

∠A+∠B+∠C=180°
And ∠ a + ∠ C = 2 ∠ B
therefore
3∠B=180°
∠B=60°.

If we know that in the triangle ABC, angle a = 2, angle B = 2 angle c, then the degree of angle a is ()

A=2B=2C
A+B+C=180
So a + 1 / 2 A + 1 / 2 a = 180
That is, a = 90

In △, the opposite sides of angles a, B and C are a, B and C respectively. Given that a + B = 5, C = radical 7, and 4sin square (a + b) / 2-cos2c = 7 / 2, find the size of angle c and the area of △ ABC

1)4sin^2(A+B)/2-cos2C=7/22（1-cos(A+B)）-cos2C=7/22-2cos(π-C)-cos2C=7/22+2cos(C)-cos2C=7/22+2cosC-2(cosC)^2+1=7/2(cosC-1)^2=0C=arcos1/2=602)cosC=1/2=(a^2+b^2-c^2)/2abab=a^2+b^2-c^2=(a+b)^2-c^2-2ab3ab...