In a triangle abc anvle ais twice as large as angle B and angle b is 20 kore than angle tge sum of interior angle of triangle is 180⁰
Let's denote angle A as x, angle B as y, and angle C as z.
Given that angle A is twice as large as angle B, we have the equation:
x = 2y
Also, angle B is 20 degrees more than the sum of angles A and C:
y = x + z + 20
We also know that the sum of the interior angles of a triangle is 180 degrees:
x + y + z = 180
Using the first equation, we can substitute x in terms of y in the second equation:
2y = y + z + 20
y = z + 20
Now we have two equations with two unknowns:
x = 2y
y = z + 20
Substituting these equations into the sum of the interior angles equation:
2y + y + z = 180
3y + z = 180
We can substitute y = z + 20 into this equation:
3(z + 20) + z = 180
4z + 60 = 180
4z = 120
z = 30
Substituting z = 30 into y = z + 20:
y = 30 + 20
y = 50
Finally, using x = 2y:
x = 2(50)
x = 100
So the triangle has angles A = 100, B = 50, and C = 30.