# The angles of a triangle are consecutive even integers. Find the measure of each angle.

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To answer this question, remember that the angles of a triangle always sum to 180 degrees. Let the three angles in question be labeled A, B, and C. Also, let A = n;

A = n; B = n + 2; C = n + 4; i.e. consecutive even numbers.

A + B + C = n + n + 2 + n + 4 = 180

3n + 6 = 180

n = 174/3 = 58

Thus,

**A = 58,**

**B= 60,**

**C = 62**

Let the smallest angle = x

Then the second angle = x+2

The third angle = x+4

But we know that the sum of the angles of a traingle = 180

Then,

x + (x+2) + (x+4) = 180

==> 3x + 6 = 180

==> 3x = 180- 6 = 174

Now divide by 3:

==> x= 174/3 = 58

Then the angles of the triangle are:

58 , 60, and 62.

Because the angles are all even, we'll note the smallest angle as 2k. The next angle is 2k+2 and the third angle is 2k+4.

Based on the fact that the sum of any 3 angles of a triangle is 180 degrees (or pi, in radians), we'll add the 3 angles:

2k + (2k+2) + (2k+4) = 180

We'll remove the brackets:

2k + 2k + 2k + 2 + 4 =180

We'll combine like terms:

3*2k + 6 = 180

6k + 6 = 180

We'll factorize by 6:

6*(k+1) = 180

We'll divide by 6:

k+1 = 180/6

k+1 = 30

We'll subtract 1 both sides:

k = 30-1

k = 29

So, the smallest angle is:

2k = 2*29

**2k = 58**

The next angle is:

**2k+2 = 58+2 = 60**

The third angle is:

2k+4 = 58+4

**2k+4 = 62**

We'll add the 3 angles to check if their sum is 180 degrees:

**58 + 60 + 62 = 180 q.e.d.**

Let the angle of the triangle which is consecutive even integers be x , x+2 and x+4 .

Then the sum of the angles : x+x+2+x+4 should be equal to 180 as sum of 3 angles of any triangle is 180 degrees.

x+(x+2)+(x+4 )= 180

3x+6 = 180

3x = 180-6 = 174

x = 174/3 = 58.

So the angles of the triangle are x = 58 deg, x+2= 58+2 = 60 deg and x+4 = 58+4 = 62 degrees.