22. In $\triangle \mathrm{PQR}, \angle \mathrm{P}=42^{\circ}, \angle \mathrm{R}=90^{\circ}$, and $\mathrm{QR}=17 \mathrm{~cm}$. Find the length of $\mathrm{PQ}$, to the nearest tenth of a centimetre. a. $25.4 \mathrm{~cm}$ c. $\quad 22.9 \mathrm{~cm}$ b. $12.6 \mathrm{~cm}$ d. $\quad 11.4 \mathrm{~cm}$ 23. In $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}, \angle \mathrm{B}=67^{\circ}$, and $\mathrm{BC}=28 \mathrm{~mm}$. Find the length of $\mathrm{AB}$, to the nearest tenth of a millimetre. a. $25.8 \mathrm{~mm}$ c. $\quad 71.7 \mathrm{~mm}$ b. $\quad 10.9 \mathrm{~mm}$ d. $\quad 30.4 \mathrm{~mm}$

Step 1 ：Given triangle PQR with angle P = 42 degrees, angle R = 90 degrees, and side QR = 17 cm. We need to find the length of PQ.

Step 2 ：Since angle R is 90 degrees, we can find angle Q using the formula: angle Q = 180 - angle P - angle R = 180 - 42 - 90 = 48 degrees.

Step 3 ：Using the sine rule, we can find the length of PQ: \(\frac{PQ}{\sin{42}} = \frac{17}{\sin{90}}\)

Step 4 ：Solving for PQ, we get: PQ = 17 * \(\frac{\sin{42}}{\sin{90}}\) = 15.30686875306328 cm

Step 5 ：Rounding to the nearest tenth, we get: \(\boxed{15.3 cm}\)

Step 6 ：Given triangle ABC with angle A = 90 degrees, angle B = 67 degrees, and side BC = 28 mm. We need to find the length of AB.

Step 7 ：Since angle A is 90 degrees, we can find angle C using the formula: angle C = 180 - angle A - angle B = 180 - 90 - 67 = 23 degrees.

Step 8 ：Using the sine rule, we can find the length of AB: \(\frac{AB}{\sin{67}} = \frac{28}{\sin{23}}\)

Step 9 ：Solving for AB, we get: AB = 28 * \(\frac{\sin{67}}{\sin{23}}\) = 65.96386624306508 mm

Step 10 ：Rounding to the nearest tenth, we get: \(\boxed{66.0 mm}\)