Here is how to solve these problems systematically:

32 (a)

Given:

• $PR = 12 \, \text{cm}$,
• $\angle R = 40^\circ$,
• Area = $32 \, \text{cm}^2$.

Find: $QR$.

Using the formula for the area of a triangle: $\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin C$ where $a$ and $b$ are two sides, and $C$ is the included angle.

Here, $PR$ acts as $b$, and $QR$ is $a$. Therefore: $32 = \frac{1}{2} \cdot 12 \cdot QR \cdot \sin(40^\circ)$

Rearranging for $QR$: $QR = \frac{2 \cdot 32}{12 \cdot \sin(40^\circ)}$

32 (b)

Given:

• $QR = 8 \, \text{mm}$,
• $\angle R = 35^\circ$,
• Area = $12 \, \text{mm}^2$.

Find: $PR$.

Using the same formula: $12 = \frac{1}{2} \cdot 8 \cdot PR \cdot \sin(35^\circ)$

Rearranging for $PR$: $PR = \frac{2 \cdot 12}{8 \cdot \sin(35^\circ)}$

33 (a)

Given:

• $a = 5.2 \, \text{cm}$,
• $\angle B = 55^\circ$,
• Area = $15 \, \text{cm}^2$.

Find: $c$.

Using the same formula: $15 = \frac{1}{2} \cdot 5.2 \cdot c \cdot \sin(55^\circ)$

Rearranging for $c$: $c = \frac{2 \cdot 15}{5.2 \cdot \sin(55^\circ)}$

33 (b)

Given:

• $a = 6.8 \, \text{cm}$,
• $\angle B = 100^\circ$,
• Area = $28 \, \text{mm}^2$.

Find: $c$.

Using the same formula: $28 = \frac{1}{2} \cdot 6.8 \cdot c \cdot \sin(100^\circ)$

Rearranging for $c$: $c = \frac{2 \cdot 28}{6.8 \cdot \sin(100^\circ)}$

If you'd like, I can calculate each value for you. Let me know if you want detailed computations or clarifications.