Solution:
Given:
The angles of a triangle are as shown in the sketch below;
Consecutive even integers are the set of integers such that each integer in the set differs from the previous integer by a difference of 2 and each integer is divisible by 2.
For the three angles to be consecutive even integers, then
[tex]\begin{gathered} x,y,z\text{ are consecutive even integers.} \\ \text{Hence,} \\ y-x=2 \\ y=x+2\ldots\ldots\ldots\ldots\ldots(1) \\ \\ \text{Also,} \\ z-y=2 \\ z=y+2 \\ z=x+2+2 \\ z=x+4\ldots.\ldots\ldots\ldots\ldots\ldots.\mathrm{}(2) \end{gathered}[/tex]Since the three angles are in the triangle, then;
[tex]\begin{gathered} x+y+z=180^0\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.(the sum of angles in a triangle)} \\ \\ \\ \text{Substituting equation (1) and (2) into the equation above,} \\ x+(x+2)+(x+4)=180^0 \\ \text{Collecting the like terms,} \\ x+x+x+2+4=180^0 \\ 3x+6=180^0 \\ 3x=180-6 \\ 3x=174 \\ \text{Dividing both sides by 3,} \\ x=\frac{174}{3} \\ x=58^0 \end{gathered}[/tex]Substituting the value of x into equations (1) and (2) to get the values of y and z,
[tex]\begin{gathered} y=x+2 \\ y=58+2 \\ y=60^0 \\ \\ \text{Also,} \\ z=x+4 \\ z=58+4 \\ z=62^0 \end{gathered}[/tex]Therefore, the measure of each angle if the angles are consecutive even integers are;
[tex]58^0,60^0,62^0[/tex]
