1. **State the problem:** We need to find the length $d$ in the smaller triangle, given angles and a side length in the larger triangle. 2. **Identify known values:** In the larger triangle, the base is 22 mm, and the angles at the base are $29^\circ$ and $103^\circ$. The smaller triangle shares side $d$ and has angles $98^\circ$ and $49^\circ$. 3. **Find the third angle in the larger triangle:** The sum of angles in a triangle is $180^\circ$. $$ 180^\circ - 29^\circ - 103^\circ = 48^\circ $$ 4. **Use the Law of Sines in the larger triangle to find the side opposite $48^\circ$ (the side adjacent to $d$):** Law of Sines: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$ Let the side opposite $29^\circ$ be 22 mm, so: $$ \frac{22}{\sin 29^\circ} = \frac{x}{\sin 48^\circ} $$ Solve for $x$: $$ x = \frac{22 \times \sin 48^\circ}{\sin 29^\circ} $$ Calculate: $$ x = \frac{22 \times 0.7431}{0.4848} = \frac{16.3482}{0.4848} = 33.72 \text{ mm} $$ 5. **Now, in the smaller triangle, use Law of Sines to find $d$:** The smaller triangle has angles $98^\circ$, $49^\circ$, and the third angle: $$ 180^\circ - 98^\circ - 49^\circ = 33^\circ $$ Side $x = 33.72$ mm is opposite $33^\circ$, and $d$ is opposite $98^\circ$. Apply Law of Sines: $$ \frac{d}{\sin 98^\circ} = \frac{33.72}{\sin 33^\circ} $$ Solve for $d$: $$ d = \frac{33.72 \times \sin 98^\circ}{\sin 33^\circ} $$ Calculate: $$ d = \frac{33.72 \times 0.9903}{0.5446} = \frac{33.38}{0.5446} = 61.3 \text{ mm} $$ 6. **Final answer:** $$ \boxed{d = 61.3 \text{ mm}} $$ (to 3 significant figures)