Is the converse of a false conditional always true as in the Truth Table?

Accroding to the Truth Table,

If $p$ is TRUE, and $q$ is FALSE, then $p\implies q$ is FALSE.
And the converse, $q \implies p$, is TRUE.

If the conditional statement is

"If two angles are congruent, they are not equal." this is a FALSE statement.

Its converse is "If two angles are not equal, they are congruent." The converse is also FALSE.

Why does this example contradict the Truth Table??

• The opposite of "If A then B" isn't "If B then A", it is "A and not B". – DanielV Aug 21 '15 at 1:06

The source of the confusion is that you are not dealing merely with implications in propositional logic, where, as you said, the truth tables ensure that $(p\implies q)\lor(q\implies p)$ is always true. Your statements about angles are universally quantified statements, even though the English language lets you hide the quantifiers. Your first statement really means "For every two angles $x$ and $y$, if $x$ and $y$ are congruent then $x$ and $y$ are not equal." Similarly for the second statement.
So the logical form of these statements is $(\forall x)(\forall y)\,(P(x,y)\implies Q(x,y))$ and $(\forall x)(\forall y)\,(Q(x,y)\implies P(x,y))$. Because of the quantifiers, it is entirely possible for both of these to be false. All that's needed for that to happen is that some particular $x_0$ and $y_0$ satisfy $P$ but not $Q$, while a different pair $x_1$ and $y_1$ satisfy $Q$ but not $P$.
It doesn't. In your example let $q =$"two angles are not equal" be false so that means they are equal. Hence the statement "If two angles are not equal, they are congruent" is vacuously true.