SOLUTION
Consider the image given below.
From the image above,

To find the measures of each angle, we need to obtain the value of small x in the diagram.
Applying the rule: Sum of angle in a triangle is 180 degrees

Then substitute the given expression, we have

Collect like terms and add

Divide both sides by 6, we have

hence, x=30
Then substitute the value of x to obtain the measure of each angles.

Since the triangle is an issoceles triangle, then

Hence

Therefore
Answer : ∠V= 28°, ∠W=28°, ∠X=124°