1. **State the problem:** We need to find the angle $\angle MLN$ in triangle $MLN$ where $ML=20$ cm, $MN=15$ cm, and $\angle M=62^\circ$. 2. **Identify known values:** - Side $ML = 20$ cm - Side $MN = 15$ cm - Angle $\angle M = 62^\circ$ 3. **Use the Law of Sines:** The Law of Sines states: $$\frac{\sin \angle MLN}{MN} = \frac{\sin \angle M}{LN} = \frac{\sin \angle LNM}{ML}$$ We want to find $\angle MLN$, so we use: $$\frac{\sin \angle MLN}{MN} = \frac{\sin \angle M}{LN}$$ But we do not know $LN$ or $\angle LNM$ yet. Instead, use the Law of Cosines to find $LN$ first. 4. **Find side $LN$ using Law of Cosines:** $$LN^2 = ML^2 + MN^2 - 2 \times ML \times MN \times \cos \angle M$$ $$LN^2 = 20^2 + 15^2 - 2 \times 20 \times 15 \times \cos 62^\circ$$ Calculate: $$LN^2 = 400 + 225 - 600 \times \cos 62^\circ$$ $$\cos 62^\circ \approx 0.4695$$ $$LN^2 = 625 - 600 \times 0.4695 = 625 - 281.7 = 343.3$$ $$LN = \sqrt{343.3} \approx 18.52 \text{ cm}$$ 5. **Use Law of Sines to find $\angle MLN$:** $$\frac{\sin \angle MLN}{MN} = \frac{\sin \angle M}{LN}$$ $$\sin \angle MLN = \frac{MN}{LN} \times \sin \angle M = \frac{15}{18.52} \times \sin 62^\circ$$ Calculate: $$\sin 62^\circ \approx 0.8829$$ $$\sin \angle MLN = 0.810 \times 0.8829 = 0.715$$ 6. **Find $\angle MLN$:** $$\angle MLN = \sin^{-1}(0.715) \approx 45.6^\circ$$ **Final answer:** $$\boxed{\angle MLN \approx 45.6^\circ}$$