1. **Problem statement:** We have an isosceles triangle $\triangle ABC$ with $AB = CB$, area $= 30$ cm$^2$, and base $AC = 12$ cm. We need to find the measure of $\angle ABC$ to the nearest degree. 2. **Known formulas and rules:** - Area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ - In an isosceles triangle with equal sides $AB = CB$, the altitude from $B$ to $AC$ bisects $AC$. 3. **Find the height:** Since $AC = 12$ cm, the altitude from $B$ bisects $AC$ into two segments of length $6$ cm each. Using the area formula: $$30 = \frac{1}{2} \times 12 \times h \implies 30 = 6h \implies h = \frac{30}{6} = 5 \text{ cm}$$ 4. **Find the length of the equal sides $AB = CB$:** Using the right triangle formed by the altitude: $$AB = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61}$$ 5. **Find $\angle ABC$:** Since $AB = CB$, $\angle ABC$ is the vertex angle opposite base $AC$. Using the cosine rule on $\triangle ABC$: $$\cos(\angle ABC) = \frac{AB^2 + CB^2 - AC^2}{2 \times AB \times CB} = \frac{61 + 61 - 144}{2 \times \sqrt{61} \times \sqrt{61}} = \frac{122 - 144}{2 \times 61} = \frac{-22}{122} = -\frac{11}{61}$$ 6. **Calculate the angle:** $$\angle ABC = \cos^{-1}\left(-\frac{11}{61}\right) \approx 100^\circ$$ **Final answer:** $\boxed{100^\circ}$ (Option A)