Question

in triangle mno, mn = 20 units, no = 15 units, and mo = 18 units. what is the approximate difference between the largest and smallest angle measures?
○ 2°
○ 5°
○ 14°
○ 28°

Explanation:

Step1: Identify largest and smallest sides

In triangle \( MNO \), the sides are \( MN = 20 \), \( NO = 15 \), \( MO = 18 \). The largest side is \( MN = 20 \) (opposite angle \( O \)), the smallest side is \( NO = 15 \) (opposite angle \( M \)).

Step2: Use the Law of Cosines for largest angle (angle \( O \))

Law of Cosines: \( \cos O=\frac{NO^{2}+MO^{2}-MN^{2}}{2\cdot NO\cdot MO} \)
Substitute values: \( NO = 15 \), \( MO = 18 \), \( MN = 20 \)
\( \cos O=\frac{15^{2}+18^{2}-20^{2}}{2\times15\times18}=\frac{225 + 324 - 400}{540}=\frac{149}{540}\approx0.2759 \)
\( O=\arccos(0.2759)\approx74.1^{\circ} \)

Step3: Use the Law of Cosines for smallest angle (angle \( M \))

Law of Cosines: \( \cos M=\frac{MN^{2}+MO^{2}-NO^{2}}{2\cdot MN\cdot MO} \)
Substitute values: \( MN = 20 \), \( MO = 18 \), \( NO = 15 \)
\( \cos M=\frac{20^{2}+18^{2}-15^{2}}{2\times20\times18}=\frac{400 + 324 - 225}{720}=\frac{499}{720}\approx0.6931 \)
\( M=\arccos(0.6931)\approx46.1^{\circ} \)

Step4: Calculate the difference

Difference \( = 74.1^{\circ}-46.1^{\circ}=28^{\circ} \) (approximate, matches the option)

Answer:

28° (corresponding to the option with 28°)