The angles of a triangle are in arithmetic progression. Prove that the triangle has the angle of 60 deg.
Since the angles are in arith metic progression.
So the angles are a ,(a+d) and (a+2d).
Then sum of the angles: a+a+d+a+2d = 3a +3d which should be 180 degree in atriangle.
Therefore 3(a+d) = 180.
Therefore a+d = 180/3 = 60.
But a+d is one of the angle in AP we assumed.
The angles of a triangle add up to 180 degree.
Now here it is given that the angles form an AP. Let the smallest angle be the first term a. If the common difference is d, the second largest angle is a+d and the largest angle is a + 2d.
So the sum of the three angles add up to 180 degree.
=> a + a+d +a+2d = 180 degree
=> 3a + 3d = 180 degree
=> a+d = 180 / 3= 60 degree.
So the second largest angle is of 60 degree.
We know that the sum of measures of the angles of a triangle is 180 degrees.
We'll note the angles of the triangle as a,b,c.
We also know that the angles of the triangle are the terms of an a.p.
So, we can write the angles as:
b = a + d
c = b + d
d is the common difference of the arithmetic progression.
The sum of the angles is 180 degrees.
a + b + c = 180
We'll substitute b and c by the relations above.
a + a + d + a + d + d = 180
We'll combine like terms:
3a + 3d = 180 degrees.
We'll factorize by 3:
3(a + d) = 180
We'll divide by 3:
a + d = 60
But a + d is the measure of the angle b, so the angle b is of 60 degrees.
a + d = b = 60 degrees