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 $P$.
If quantifiers are not involved, for example if you have two particular angles and are making statements about just this pair, not angles in general, then a proposition and its converse will not both be false.
I would like to show you this in truth tables, hopefully it resolves some of the confusion. Of course, you should read the other answers as well, Patrick Stevens answer I like personally.
The connective "$\Rightarrow$" is defined according to this truth table:
$$
\begin{array}{c|l|c|}
P & \text{Q} & \text{P $\Rightarrow$ Q} \\
\hline
T & T & T \\
T & F & F \\
F & T & T \\
F & F & T
\end{array}
$$
This should be in your book. As you can see, $P \Rightarrow Q$ is false only when $P$ is true and $Q$ is false, otherwise $P \Rightarrow Q$ is true.
For the contrapositive, we can draw the following table:
$$
\begin{array}{c|l|c|c}
P & Q & \neg Q & \neg P & \neg Q \Rightarrow \neg P \\
\hline
T & T & F & F & T \\
T & F & T & F & F\\
F & T & F & T & T\\
F & F & T & T & T
\end{array}
$$
As you can see, for any valuation of $P, Q$ (for any assignment of $T,F$ to $P,Q$), whenever $ P \Rightarrow Q$ is true, so is $\neg Q \Rightarrow \neg P $.
So as you can see, if $\neg Q$ is false, and $\neg P$ is true, then $\neg Q \Rightarrow \neg P$ is true, by the third line in the above table. This is exactly in line with what is says in your book.
For the converse (or inverse) , we have:
$$
\begin{array}{c|l|c|c}
P & \text{Q} & \text{P $\Rightarrow$ Q} & Q \Rightarrow P \\
\hline
T & T & T & T \\
T & F & F & T \\
F & T & T & F \\
F & F & T & T
\end{array}
$$
As you can see, they don't match for all valuations, in particular, if $P$ is true, and $Q$ is false, then one implication is true and the other isn't.
Finally: if $A$ is the statement "you are in Paris" and $B$ is the statement "you are in France" and you know $A \Rightarrow B$ is true, then if you are not in France, then you are not in Paris ($\neg B \Rightarrow \neg A$).
And of course, if you are not in Paris ($\neg A$ is true) then you could be on Mars for all we know. You are not necessarily in France.
EDIT: I would like to encourage you to look at more mathematical examples, similar to other peoples' examples in their answers, natural language is full of vague statements that are not always as clear cut as mathematical ones.
Best Answer
Let's agree that 'if you are in Paris, then you are in France' ($A \implies B$). (We could get picky, and say maybe you're in Paris, Texas; but let's not!).
But then the converse ($B \implies A$) is not automatically true: for example, we can't then deduce from 'if you are in Paris, then you are in France' that 'if you are in France, then you are in Paris'. In that sense you're right that 'it's not logical' to say the converse is always true.
Now, bear in mind that the converse might be true, or it might not. It must be proven either way based on other information. And it turns out that we know enough other things, about European geography in this case, that we can also prove that the converse is not true.
On the other hand, what we can always deduce is called the contrapositive: once we accept the truth of 'if you are in Paris, then you are in France', then we always automatically can say 'if you are not in France, then you are not in Paris' ($\neg B \implies \neg A$). That will always be true (at least, in the world of mathematical language).