1. **State the problem:** We have an isosceles triangle with two equal sides each measuring 11.52 inches and a vertex angle of 32° 15'. We need to find the length of the base. 2. **Formula and approach:** In an isosceles triangle, the base can be found by splitting the triangle into two right triangles by drawing an altitude from the vertex angle to the base. This altitude bisects the base and the vertex angle. The base length $b$ is twice the length of the half-base. Using the law of cosines or trigonometry, the half-base $\frac{b}{2}$ can be found by: $$\frac{b}{2} = 11.52 \times \sin\left(\frac{32^\circ 15'}{2}\right)$$ 3. **Convert the vertex angle to decimal degrees:** $$32^\circ 15' = 32 + \frac{15}{60} = 32.25^\circ$$ 4. **Calculate half the vertex angle:** $$\frac{32.25^\circ}{2} = 16.125^\circ$$ 5. **Calculate half the base:** $$\frac{b}{2} = 11.52 \times \sin(16.125^\circ)$$ Using a calculator: $$\sin(16.125^\circ) \approx 0.277$$ So: $$\frac{b}{2} = 11.52 \times 0.277 = 3.19$$ 6. **Calculate the full base:** $$b = 2 \times 3.19 = 6.38$$ **Answer:** The base of the isosceles triangle is approximately **6.38 inches**.