1. **Problem statement:** We have an isosceles triangle with two equal sides of length $n$ and a vertical side of length 4.2 mm. One of the base angles is $52^\circ$. We need to find the length $n$ to 1 decimal place. 2. **Understanding the triangle:** In an isosceles triangle, the two equal sides are opposite the equal angles. Here, the vertical side (4.2 mm) is the base, and the two equal sides $n$ meet at the top vertex. 3. **Using the Law of Cosines:** The Law of Cosines relates the sides and angles of any triangle: $$ c^2 = a^2 + b^2 - 2ab \cos(C) $$ where $c$ is the side opposite angle $C$. 4. **Assigning values:** Let the base be $c = 4.2$ mm, and the equal sides be $a = b = n$. The angle opposite the base is the vertex angle, which is $180^\circ - 2 \times 52^\circ = 76^\circ$. 5. **Apply Law of Cosines:** $$ 4.2^2 = n^2 + n^2 - 2 \times n \times n \times \cos(76^\circ) $$ which simplifies to $$ 4.2^2 = 2n^2 - 2n^2 \cos(76^\circ) $$ 6. **Simplify and solve for $n^2$:** $$ 4.2^2 = 2n^2 (1 - \cos(76^\circ)) $$ $$ n^2 = \frac{4.2^2}{2(1 - \cos(76^\circ))} $$ 7. **Calculate numeric values:** $$ 4.2^2 = 17.64 $$ $$ \cos(76^\circ) \approx 0.2419 $$ $$ 1 - 0.2419 = 0.7581 $$ 8. **Calculate $n^2$:** $$ n^2 = \frac{17.64}{2 \times 0.7581} = \frac{17.64}{1.5162} \approx 11.63 $$ 9. **Find $n$:** $$ n = \sqrt{11.63} \approx 3.41 $$ 10. **Final answer rounded to 1 decimal place:** $$ n \approx 3.4 \text{ mm} $$