1. **State the problem:** We are given a triangle with vertices A, B, and C. Given: - Side $b = AC = 4$ - Side $c = AB = 6$ - Angle $A = 36^\circ$ We need to find the length of side $a = BC$. 2. **Formula used:** We can use the Law of Cosines to find side $a$ when two sides and the included angle are known: $$a^2 = b^2 + c^2 - 2bc \cos A$$ 3. **Apply the formula:** Substitute the known values: $$a^2 = 4^2 + 6^2 - 2 \times 4 \times 6 \times \cos 36^\circ$$ 4. **Calculate each term:** $$a^2 = 16 + 36 - 48 \cos 36^\circ$$ 5. **Evaluate $\cos 36^\circ$:** $$\cos 36^\circ \approx 0.8090$$ 6. **Substitute and simplify:** $$a^2 = 52 - 48 \times 0.8090 = 52 - 38.832 = 13.168$$ 7. **Find $a$ by taking the square root:** $$a = \sqrt{13.168} \approx 3.63$$ **Final answer:** The length of side $a$ is approximately **3.63** units.