1. **State the problem:** We are given a triangle LMN with area 133 cm², side LM = 16 cm, and side MN = 21 cm. We need to find the angles $\angle LMN$ and $\angle MNL$. 2. **Recall the formula for the area of a triangle using two sides and the included angle:** $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are two sides and $C$ is the angle between them. 3. **Identify the sides and angle for the area formula:** The area is given by sides LM and MN with the included angle $\angle LMN$. So, $$133 = \frac{1}{2} \times 16 \times 21 \times \sin(\angle LMN)$$ 4. **Solve for $\sin(\angle LMN)$:** $$133 = 168 \times \frac{\sin(\angle LMN)}{2} = 84 \sin(\angle LMN)$$ $$\sin(\angle LMN) = \frac{133}{84} \approx 1.5833$$ Since $\sin(\theta)$ cannot be greater than 1, this suggests the angle $\angle LMN$ is not between sides LM and MN. We need to reconsider which angle is between LM and MN. 5. **Check the triangle sides and angles:** - Side LM connects points L and M. - Side MN connects points M and N. - The angle between LM and MN is at point M, which is $\angle LMN$. So the angle between LM and MN is indeed $\angle LMN$. 6. **Since $\sin(\angle LMN) > 1$ is impossible, the given data might mean the area is formed by sides LN and MN or LM and LN. Let's find side LN using the Law of Cosines or check if the problem expects us to find angles using Law of Cosines.** 7. **Use the formula for area with sides LM and LN or MN and LN. But we don't have LN. Let's find LN using the Law of Cosines if we find one angle first. Alternatively, use the formula for area with sides LM and MN and angle between them, but since $\sin(\angle LMN) > 1$, the angle is not between LM and MN.** 8. **Try to find side LN using the formula for area with sides LM and MN and angle between them, but since it failed, try to find the height from N to LM:** Area = $\frac{1}{2} \times base \times height$ Using LM as base: $$133 = \frac{1}{2} \times 16 \times h$$ $$h = \frac{2 \times 133}{16} = 16.625$$ 9. **Use Pythagoras theorem in triangle with base LM = 16, height h = 16.625, and side MN = 21 to find angle $\angle LMN$:** Drop perpendicular from N to LM, call foot P. Then, $$NP = h = 16.625$$ $$LP = x$$ $$MP = 16 - x$$ Using side MN = 21: $$MN^2 = NP^2 + (MP)^2$$ $$21^2 = 16.625^2 + (16 - x)^2$$ Calculate: $$441 = 276.39 + (16 - x)^2$$ $$(16 - x)^2 = 441 - 276.39 = 164.61$$ $$16 - x = \sqrt{164.61} \approx 12.83$$ $$x = 16 - 12.83 = 3.17$$ 10. **Find $\angle LMN$ using trigonometry:** $\angle LMN$ is angle at M between LM and MN. Using right triangle with sides: $$\tan(\angle LMN) = \frac{NP}{MP} = \frac{16.625}{12.83} \approx 1.296$$ $$\angle LMN = \arctan(1.296) \approx 52.5^\circ$$ 11. **Find $\angle MNL$ using Law of Cosines:** We know sides LM = 16, MN = 21, and height from N to LM is 16.625. Find side LN using Pythagoras: $$LN^2 = NP^2 + LP^2 = 16.625^2 + 3.17^2 = 276.39 + 10.05 = 286.44$$ $$LN = \sqrt{286.44} \approx 16.92$$ Use Law of Cosines at N to find $\angle MNL$: $$\cos(\angle MNL) = \frac{MN^2 + LN^2 - LM^2}{2 \times MN \times LN} = \frac{21^2 + 16.92^2 - 16^2}{2 \times 21 \times 16.92}$$ Calculate numerator: $$441 + 286.44 - 256 = 471.44$$ Calculate denominator: $$2 \times 21 \times 16.92 = 710.64$$ $$\cos(\angle MNL) = \frac{471.44}{710.64} \approx 0.663$$ $$\angle MNL = \arccos(0.663) \approx 48.5^\circ$$ 12. **Final answers:** $$\angle LMN \approx 52.5^\circ$$ $$\angle MNL \approx 48.5^\circ$$