13.59 In a right triangle $ABC$, the hypotenuse $c=5$, and the two legs $a \leqslant 3, b \geqslant 3$, then the maximum value of $a+b$ is 
(A) $5 \sqrt{2}$.
(B) 7 .
(C) $4 \sqrt{3}$.
(D) 6 .
(China Hubei Province Huanggang Region Junior High School Mathematics Competition, 1993)
[Solution 1] As shown, $a+b=5 \sin A + 5 \cos A$,
and
$$
\begin{array}{l}
\sin A + \cos A = \sqrt{2} \sin \left(45^{\circ} + A\right), \\
(\sin A + \cos A)_{\max} = \sqrt{2},
\end{array}
$$

we know that
but at this point $A = 45^{\circ}$, then $a = b = \frac{5}{2} \sqrt{2}$, which contradicts $a \leqslant 3$.
It is easy to see: the sum of the lengths of the chords connecting point $C$ on the semicircle with diameter $AB$ to $A$ and $B$ is maximized when the absolute difference between $\angle CBA$ and $\angle CAB$ is minimized (this can also be seen from the $\circledast$ formula, noting that $\sin x$ is an increasing function on $\left(0, \frac{\pi}{2}\right)$), or when the difference between the two chords is minimized.

Since $a$ is maximized at 3, let $a = 3$, then $b = \sqrt{5^{2} - 3^{2}} = 4$, at this point $a + b$ is maximized.
Thus, $(a + b)_{\max} = 7$.
Therefore, the answer is $(B)$.
[Solution 2] Given $a^{2} + b^{2} = 25$, then
$$
y = a + b = a + \sqrt{25 - a^{2}},
$$

which means
$$
2a^{2} - 2ay + y^{2} - 25 = 0.
$$

Since $\because a \leqslant 3 \leqslant b$,
$\therefore \quad a = \frac{2y - \sqrt{4y^{2} - 8(y^{2} - 25)}}{2 \cdot 2} \leqslant 3$,
$\therefore \quad (y - 6)^{2} \leqslant 50 - y^{2}$, which means $y \leqslant 7$.
Thus, $y_{\max} = 7$. Therefore, $(a + b)_{\max} = 7$.
Hence, the answer is $(B)$.