Name:
ID: A
44. Martina's toy boat has a base that is $50 \mathrm{~cm}$ long. The width of the sail is half the width of the base of the boat. The sail of the boat is represented by $\triangle \mathrm{MNO}$ in the diagram. $\angle \mathrm{N}$ is a right angle and $\angle \mathrm{O}$ is $50^{\circ}$. Find the height, MN, of the sail, to the nearest centimetre.
Base of Boat

Step 1 ：Given the base of the boat is 50 cm, the width of the sail (NO) is half the width of the base, so NO = \(\frac{1}{2} \times 50 = 25 \mathrm{~cm}\).

Step 2 ：In right triangle MNO, we have angle O = 50 degrees, and we need to find the length of side MN. We can use the tangent function to relate the angle O, the length of side MN, and the width of the sail (NO): \(\tan(O) = \frac{MN}{NO}\).

Step 3 ：Plug in the values: \(\tan(50) = \frac{MN}{25}\).

Step 4 ：Solve for MN: \(MN = 25 \times \tan(50) \approx 29.79\).

Step 5 ：Round the height of the sail, MN, to the nearest centimeter: \(\boxed{30 \mathrm{~cm}}\).