1. **Problem statement:** We have a right triangle ABC with a right angle at C. The angle $\theta$ is at vertex A. Side BC is opposite $\theta$, and side AC is adjacent to $\theta$. Given that BC is 20% longer than AC, find $\theta$ to 1 decimal place. 2. **Formula and rules:** In a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC}$$ 3. **Express the ratio:** Since BC is 20% longer than AC, we write: $$BC = AC + 0.2 \times AC = 1.2 \times AC$$ 4. **Calculate tangent:** Substitute into the tangent formula: $$\tan(\theta) = \frac{BC}{AC} = \frac{1.2 \times AC}{AC}$$ 5. **Simplify the fraction:** Cancel $AC$ in numerator and denominator: $$\tan(\theta) = \frac{\cancel{1.2 \times AC}}{\cancel{AC}} = 1.2$$ 6. **Find $\theta$:** Use the inverse tangent function: $$\theta = \tan^{-1}(1.2)$$ 7. **Calculate the angle:** Using a calculator, $$\theta \approx 50.2^\circ$$ **Final answer:** $$\boxed{\theta = 50.2^\circ}$$