Question 53, Let the area of $\triangle A B C$ be $1, \angle A$ be opposite to the side length $a$, try to find the minimum value of $\mathrm{a}^{2}+\frac{1}{\sin \mathrm{A}}$.

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Question 53, Solution: Let the other two sides of $\triangle A B C$ be $b$ and $c$, respectively. According to the given conditions, we have $b c=\frac{2}{\sin A}$. Thus,
$$
\begin{array}{l}
a^{2}+\frac{1}{\sin A}=b^{2}+c^{2}-2 b c \cdot \cos A+\frac{1}{\sin A} \geq 2 b c-2 b c \cdot \cos A+\frac{1}{\sin A} \\
=2 b c(1-\cos A)+\frac{1}{\sin A}=\frac{4(1-\cos A)}{\sin A}+\frac{1}{\sin A}=\frac{5-4 \cos A}{\sin A} \\
=\frac{\sqrt{\left(4^{2}+3^{2}\right)\left(\cos ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~A}\right)}-4 \cos \mathrm{A}}{\sin \mathrm{A}} \geq \frac{(4 \cos \mathrm{A}+3 \sin \mathrm{A})-4 \cos \mathrm{A}}{\sin \mathrm{A}}=3 \\
\end{array}
$$

Equality holds if and only if $\left\{\begin{array}{c}b=c \\ b c=\frac{2}{\sin A} \\ \frac{\cos A}{4}=\frac{\sin A}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}b=c=\frac{\sqrt{30}}{3} \\ A=\arcsin \frac{3}{5}\end{array}\right.\right.$