The vertex angle of an isosceles triangle is 20° less than the sum of the base angles. Which system of equations can be used to find the measure of the vertex and base angles?

The base angles are equal and we'll call those "x" and "x".
We'll call the vertex angle y.
A) y + 2x = 180
B) y + 20 = 2x we can algebraically shift "B)" into
B) y - 2x = -20 then we'll add that to "A)"
A) y + 2x = 180
2y = 160
"y" the vertex angle = 80 degrees
each base angle = 50 degrees

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jajumonac jajumonac · Answer 2 · 2020-01-29T01:55:19+01:00

Answer:

The vertex angle is 80°, and each base angle is 20°.

Step-by-step explanation:

By definition, we know that the angles of the base of an isosceles triangle are equal.

"The vertex angle of an isosceles triangles is 20° less than the sum of the base angles" this expression can be expressed like

[tex]V=B+B-20\°[/tex]

Where [tex]V[/tex] is the vertex angle and [tex]B[/tex] represents each base angle.

Also, we use the theorem which states that the sum of all interior angles of a triangle is equal to 180°, that is

[tex]V+B+B=180\°[/tex]

Now, to find each angle, we replace the first expression into the second one and solve for [tex]B[/tex]

[tex]B+B-20\°+B+B=180\°\\4B=180\°+20\°\\B=\frac{200\°}{4}\\ B=50\°[/tex]

Then, we replace this value in one equation to find the other value

[tex]V=50\°+50\°-20\°=80\°[/tex]

Therefore, the vertex angle is 80°, and each base angle is 20°.