Every triangle has 180 degrees.  An isoceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x=\) vertex angle and \(\displaystyle 3x-15=\) base angle.

Then the equation to solve becomes:

\(\displaystyle x+(3x-15)+(3x-15)=180\), or \(\displaystyle 7x-30=180\).

Then the vertex angle is \(\displaystyle 30\), the base angle is \(\displaystyle 75\), and the product is \(\displaystyle 2250\).

Because the interior angles of a triangle add up to 180°, we can create an equation using the variables given in the problem: x+2x+(3x+30)=180. This simplifies to 6X+30=180. When we subtract 30 from both sides, we get 6x=150. Then, when we divide both sides by 6, we get x=25. Because Angle B=2x degrees, we multiply 25 times 2. Thus, Angle B is equal to 50°. If you got an answer of 25, you may have forgotten to multiply by 2. If you got 105, you may have found Angle C instead of Angle B.

Every triangle has 180 degrees.  An isosceles triangle has one vertex angle and two congruent base angles.

Let \(\displaystyle x\) = vertex angle and \(\displaystyle 3x + 6\) = base angle.

Then the equation to solve becomes

 \(\displaystyle x + (3x + 6) + (3x + 6) = 180\)

or

\(\displaystyle 7x + 12 = 180\).

Solving for \(\displaystyle x\) gives a vertex angle of 24 degrees and a base angle of 78 degrees.

Every triangle has \(\displaystyle 180^o\). An isosceles triangle has one vertex ange, and two congruent base angles.

Let \(\displaystyle x\) be the vertex angle and \(\displaystyle 3x + 13\) be the base angle.

The equation to solve becomes \(\displaystyle x + (3x + 13) + (3x + 13) = 180^o\), since the base angle occurs twice.

\(\displaystyle 7x + 26 = 180^o\)

\(\displaystyle 7x=154^o\)

\(\displaystyle x=22^o\)

Now we can solve for the vertex angle.

\(\displaystyle 3x+13=3(22)+13=79^o\)

The difference between the vertex angle and the base angle is \(\displaystyle 79^o - 22^o = 57^o\).

Each triangle has \(\displaystyle 180\) degrees.

An isoceles triangle has two congruent base angles and one vertex angle.

Let \(\displaystyle x=\) vertex angle and \(\displaystyle 2x-5=\) base angle.

Then the equation to solve becomes \(\displaystyle x+(2x-5)+(2x-5)=180\) or \(\displaystyle 5x-10=180\).

Add \(\displaystyle 10\) to both sides to get \(\displaystyle 5x=190\).

Divide both sides by \(\displaystyle 5\) to get \(\displaystyle x=38\) vertex angle and \(\displaystyle 2x-5=71\) base angles, so the sum of the angles is \(\displaystyle 109\).

Every triangle contains \(\displaystyle 180\) degrees.  An isoceles triangle has two congruent base angles and one vertex angle.

Let \(\displaystyle x=\) the vertex angle and \(\displaystyle 2x =\) the base angle

So the equation to solve becomes \(\displaystyle x+2x+2x=180\) or \(\displaystyle 5x=180\) and dividing by \(\displaystyle 5\) gives \(\displaystyle x = 36\) for the vertex angle and \(\displaystyle 2x = 72\) for the base angle, so the sum is \(\displaystyle 108\)