1. **State the problem:** A pilot uses triangles to find the angle of elevation $\angle A$ from the ground to her plane. Given the triangle with sides 12 km, 20 km, and 20 km, and an angle of 40° at point C, find $m\angle A$. 2. **Identify the triangle and known values:** The triangle has two sides of 20 km each and one side of 12 km. The angle at C is 40°. 3. **Use the Law of Cosines to find the side opposite $\angle A$ if needed, or use Law of Sines to find $\angle A$:** Since two sides adjacent to $\angle A$ are 20 km and 12 km, and the angle between them is 40°, we can use the Law of Cosines to find the side opposite $\angle A$ (let's call it side $a$): $$a^2 = 20^2 + 12^2 - 2 \times 20 \times 12 \times \cos 40^\circ$$ Calculate: $$a^2 = 400 + 144 - 480 \times \cos 40^\circ$$ $$a^2 = 544 - 480 \times 0.7660 = 544 - 367.68 = 176.32$$ $$a = \sqrt{176.32} \approx 13.28 \text{ km}$$ 4. **Use Law of Sines to find $\angle A$:** $$\frac{\sin \angle A}{12} = \frac{\sin 40^\circ}{13.28}$$ Multiply both sides by 12: $$\sin \angle A = \frac{12 \times \sin 40^\circ}{13.28}$$ Calculate numerator: $$12 \times 0.6428 = 7.7136$$ Divide: $$\sin \angle A = \frac{7.7136}{13.28} \approx 0.5807$$ 5. **Find $\angle A$ by taking inverse sine:** $$\angle A = \sin^{-1}(0.5807) \approx 35.5^\circ$$ **Final answer:** $$m\angle A \approx 35.5^\circ$$