1. **Problem statement:** We need to find the size of angle $ABC$ in the quadrilateral $ABCD$ where $AB=17$ mm, $BC=14$ mm, $CD=25$ mm, $DA=19$ mm, and angle $D=53^\circ$. 2. **Approach:** Since we know one angle and all side lengths, we can use the Law of Cosines and properties of triangles to find angle $ABC$. 3. **Step 1: Find diagonal $AC$ using triangle $ADC$** In triangle $ADC$, sides are $AD=19$, $DC=25$, and angle $D=53^\circ$. Using Law of Cosines: $$AC^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos(53^\circ)$$ Calculate: $$AC^2 = 19^2 + 25^2 - 2 \times 19 \times 25 \times \cos(53^\circ)$$ $$= 361 + 625 - 950 \times 0.6018$$ $$= 986 - 571.71 = 414.29$$ So, $$AC = \sqrt{414.29} \approx 20.36$$ 4. **Step 2: Find angle $ABC$ in triangle $ABC$** Triangle $ABC$ has sides $AB=17$, $BC=14$, and diagonal $AC \approx 20.36$. Using Law of Cosines to find angle $B$: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle ABC)$$ Rearranged: $$\cos(\angle ABC) = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$$ Calculate numerator: $$17^2 + 20.36^2 - 14^2 = 289 + 414.29 - 196 = 507.29$$ Calculate denominator: $$2 \times 17 \times 20.36 = 692.24$$ So, $$\cos(\angle ABC) = \frac{507.29}{692.24} \approx 0.7327$$ 5. **Step 3: Calculate angle $ABC$** $$\angle ABC = \cos^{-1}(0.7327) \approx 43.2^\circ$$ Rounded to the nearest degree: $$\boxed{43^\circ}$$