1. **Problem statement:** Given a triangle with sides 38, 33, and an angle of 35° at vertex A adjacent to side 38, find the angle A and the length of side $x$ opposite the 35° angle. 2. **Identify known values:** - Side adjacent to angle 35°: 38 - Other side: 33 - Angle at vertex A: 35° - Side opposite angle 35°: $x$ 3. **Use the Law of Cosines to find side $x$: ** The Law of Cosines states: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ Here, $c = x$, $a = 38$, $b = 33$, and $C = 35^\circ$. 4. **Calculate $x$: ** $$x^2 = 38^2 + 33^2 - 2 \times 38 \times 33 \times \cos(35^\circ)$$ Calculate each term: $$38^2 = 1444$$ $$33^2 = 1089$$ $$2 \times 38 \times 33 = 2508$$ $$\cos(35^\circ) \approx 0.8192$$ So, $$x^2 = 1444 + 1089 - 2508 \times 0.8192 = 2533 - 2053.6 = 479.4$$ 5. **Find $x$: ** $$x = \sqrt{479.4} \approx 21.89$$ 6. **Find angle A:** Since angle A is given as 35°, the problem likely asks to confirm or find the other angles. 7. **Use Law of Sines to find another angle, say angle opposite side 33:** $$\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$$ Here, $$\frac{\sin(35^\circ)}{21.89} = \frac{\sin(B)}{33}$$ Calculate: $$\sin(35^\circ) \approx 0.574$$ So, $$\frac{0.574}{21.89} = \frac{\sin(B)}{33} \Rightarrow \sin(B) = \frac{33 \times 0.574}{21.89} = 0.865$$ 8. **Find angle B:** $$B = \arcsin(0.865) \approx 59.9^\circ$$ 9. **Find angle C:** Sum of angles in triangle is 180°: $$C = 180^\circ - 35^\circ - 59.9^\circ = 85.1^\circ$$ **Final answers:** - Angle A = 35° (given) - Side $x \approx 21.89$