在下图中，三角形 ABC 全等于三角形 PNA。 已知 BC = 22 cm，AP = 32 cm 且角 ACB = 91°，计算

1. Identify the corresponding sides and included angle in triangle $PNA$.

Since $\triangle ABC \cong \triangle PNA$, the order gives
$A \leftrightarrow P$, $B \leftrightarrow N$, $C \leftrightarrow A$.

Hence
$AC = AP = 32\text{ cm}$, $AN = BC = 22\text{ cm}$, and
$\angle PAN = \angle ACB = 91^\circ$.

2. Find $PN$ using the cosine rule in $\triangle APN$.

In $\triangle APN$,
$PN^2 = AP^2 + AN^2 - 2(AP)(AN)\cos \angle PAN$
$PN^2 = 32^2 + 22^2 - 2(32)(22)\cos 91^\circ$
$PN \approx 39.148\text{ cm}$

Since $AB$ corresponds to $PN$,
$AB = PN \approx 39.148\text{ cm}.$

3. Find $BP$ and subtract $PN$ to get $BN$.

Since $B,N,P$ are collinear, $\angle APB = \angle APN$.

First find $\cos \angle APN$ from $\triangle APN$:
$\cos \angle APN = \frac{AP^2 + PN^2 - AN^2}{2(AP)(PN)}$
$\cos \angle APN = \frac{32^2 + 39.148^2 - 22^2}{2(32)(39.148)} \approx 0.8272$

In $\triangle APB$, $AB = PN \approx 39.148\text{ cm}$, so
$AB^2 = AP^2 + BP^2 - 2(AP)(BP)\cos \angle APB.$
Solving gives
$BP \approx 61.245\text{ cm}.$

Therefore
$BN = BP - PN$
$BN \approx 61.245 - 39.148 = 22.097\text{ cm}$
$BN \approx 22.1\text{ cm}.$

1. Identify the needed sides and angle in $\triangle APN$.

Since $\triangle ABC \cong \triangle PNA$, the corresponding vertices are
$A \leftrightarrow P,\quad B \leftrightarrow N,\quad C \leftrightarrow A.$
Hence
$AN = BC = 22\text{ cm}$
and
$\angle PAN = \angle ACB = 91^\circ.$
Also, $AP = 32\text{ cm}$.

2. Calculate $\angle APN$ using the cosine rule.

First find $PN$:
$PN^2=32^2+22^2-2(32)(22)\cos91^\circ$
$PN=39.15\text{ cm}$
Now use the cosine rule for $\angle APN$:
$\cos \angle APN=\frac{32^2+39.15^2-22^2}{2(32)(39.15)}$
$\angle APN=34.2^\circ\text{ }\text{(to 1 d.p.)}$