1. **State the problem:** We need to find the value of angle $x$ in quadrilateral $ABCD$ composed of two triangles $ABC$ and $BCD$ with given sides and angles. 2. **Given data:** - Triangle $ABC$: $AC=5.1$, $AB=3.4$, $\angle CAB=39^\circ$, $\angle ABC=73^\circ$ - Triangle $BCD$: $CD=3.4$, $BD=5.1$, $\angle BCD = x$ 3. **Find:** $x$ (angle $BCD$). 4. **Step 1: Find $\angle ACB$ in triangle $ABC$** Since the sum of angles in a triangle is $180^\circ$: $$\angle ACB = 180^\circ - 39^\circ - 73^\circ = 68^\circ$$ 5. **Step 2: Find length $BC$ using Law of Cosines in triangle $ABC$** Law of Cosines formula: $$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle CAB)$$ Substitute values: $$BC^2 = 3.4^2 + 5.1^2 - 2 \cdot 3.4 \cdot 5.1 \cdot \cos(39^\circ)$$ Calculate: $$3.4^2 = 11.56, \quad 5.1^2 = 26.01$$ $$\cos(39^\circ) \approx 0.7771$$ $$BC^2 = 11.56 + 26.01 - 2 \cdot 3.4 \cdot 5.1 \cdot 0.7771 = 37.57 - 26.93 = 10.64$$ $$BC = \sqrt{10.64} \approx 3.26$$ 6. **Step 3: Use Law of Cosines in triangle $BCD$ to find $x = \angle BCD$** Given sides: $$BD = 5.1, CD = 3.4, BC = 3.26$$ Apply Law of Cosines to find $x$: $$BC^2 = BD^2 + CD^2 - 2 \cdot BD \cdot CD \cdot \cos(x)$$ Rearranged for $\cos(x)$: $$\cos(x) = \frac{BD^2 + CD^2 - BC^2}{2 \cdot BD \cdot CD}$$ Substitute values: $$\cos(x) = \frac{5.1^2 + 3.4^2 - 3.26^2}{2 \cdot 5.1 \cdot 3.4} = \frac{26.01 + 11.56 - 10.64}{34.68} = \frac{26.93}{34.68} \approx 0.7765$$ 7. **Step 4: Calculate $x$** $$x = \arccos(0.7765) \approx 39.1^\circ$$ **Final answer:** $$\boxed{x \approx 39.1^\circ}$$