find the smallest angle of this pentagon x, x-15, x-50, x-30, x-40

Cannot be answered with the information given.

e.g. let x = 60
then you have sides of 60, 45, 10, 30, and 20
Use a ruler and a compass and a scale factor of 1/10,
that is, sides of 6, 4.5, 1, 3, and 2 inches (or cm or..)
construct such a pentagon.
(I actually cut some bamboo skewers to those lengths and formed a pentagon
I was able to "squish" the shape and change the angles.)
Try it.

There is no unique answer to the question.

For some interesting problems relating to this topic, try this question:
Show that a quadrilateral of sides 3, 4, 5, and 6 has the largest area when opposite angles are supplementary.

To find the smallest angle of the pentagon, we need to find the smallest value among the angles x, x-15, x-50, x-30, and x-40.

Let's start by finding the expressions for each angle:
Angle 1: x degrees
Angle 2: (x-15) degrees
Angle 3: (x-50) degrees
Angle 4: (x-30) degrees
Angle 5: (x-40) degrees

To find the smallest angle, we can compare the expressions for each angle. Let's simplify each expression:

Angle 1: x degrees
Angle 2: x - 15 degrees
Angle 3: x - 50 degrees
Angle 4: x - 30 degrees
Angle 5: x - 40 degrees

Now, we can compare the expressions:

Angle 1 < Angle 2 < Angle 3 < Angle 4 < Angle 5

Since Angle 1 has the smallest expression (x degrees), it represents the smallest angle of the pentagon.

Therefore, the smallest angle of the pentagon is x degrees.