1. **State the problem:** Given a triangle with an angle of 40° and two sides adjacent to angle A measuring 21 and 27, find the angle A and the length of side $x$ opposite the 40° angle. 2. **Identify known values:** - Angle opposite side $x$ is $40^\circ$. - Sides adjacent to angle A are 21 and 27. - Angle A is between sides 21 and 27. 3. **Find angle A using the triangle angle sum rule:** The sum of angles in a triangle is $180^\circ$. Let angle at the third vertex be $B$. 4. **Calculate angle A:** Since angle A is between sides 21 and 27, and the other known angle is $40^\circ$, we can find angle A by first finding angle B using the Law of Cosines or Law of Sines. 5. **Use Law of Cosines to find side $x$ opposite $40^\circ$:** $$x^2 = 21^2 + 27^2 - 2 \times 21 \times 27 \times \cos(40^\circ)$$ 6. **Calculate $x^2$:** $$x^2 = 441 + 729 - 2 \times 21 \times 27 \times \cos(40^\circ)$$ $$x^2 = 1170 - 1134 \times \cos(40^\circ)$$ 7. **Calculate $\cos(40^\circ)$:** $$\cos(40^\circ) \approx 0.7660$$ 8. **Substitute and simplify:** $$x^2 = 1170 - 1134 \times 0.7660 = 1170 - 868.644 = 301.356$$ 9. **Find $x$ by taking the square root:** $$x = \sqrt{301.356} \approx 17.36$$ 10. **Find angle A using Law of Sines:** $$\frac{\sin A}{21} = \frac{\sin 40^\circ}{x}$$ 11. **Substitute known values:** $$\sin A = \frac{21 \times \sin 40^\circ}{x} = \frac{21 \times 0.6428}{17.36} = \frac{13.4988}{17.36} = 0.7773$$ 12. **Calculate angle A:** $$A = \arcsin(0.7773) \approx 51.0^\circ$$ 13. **Summary:** - Angle A is approximately $51.0^\circ$. - Side $x$ is approximately $17.36$. Note: The values differ from the user's given values because the problem's description and given values suggest a different triangle configuration. The above solution follows the standard approach for the given data.