1. **Problem statement:** A parallelogram has the largest side $a=55$ cm. A diagonal makes angles $30^\circ$ and $50^\circ$ with the pair of adjacent sides. We need to find the length of this diagonal. 2. **Formula and concepts:** In a parallelogram, the diagonal length can be found using the Law of Cosines or vector addition. Here, the diagonal $d$ can be expressed as the vector sum of two adjacent sides $\vec{a}$ and $\vec{b}$. 3. Let the sides be $a=55$ cm and $b$ (unknown). The diagonal $d$ makes angles $30^\circ$ with side $a$ and $50^\circ$ with side $b$. Since the diagonal is the sum of vectors $\vec{a}$ and $\vec{b}$, the angle between $\vec{a}$ and $\vec{b}$ is $180^\circ - (30^\circ + 50^\circ) = 100^\circ$. 4. Using the Law of Cosines for the diagonal: $$ d^2 = a^2 + b^2 + 2ab \cos(100^\circ) $$ 5. Using the angle relations for the diagonal $d$: $$ d = a \cos 30^\circ + b \cos 50^\circ $$ and $$ d = a \sin 30^\circ + b \sin 50^\circ $$ These come from projecting the diagonal onto the directions of $a$ and $b$. 6. From the projections, set up the system: $$ d = 55 \cos 30^\circ + b \cos 50^\circ $$ $$ d = 55 \sin 30^\circ + b \sin 50^\circ $$ 7. Equate the two expressions for $d$: $$ 55 \cos 30^\circ + b \cos 50^\circ = 55 \sin 30^\circ + b \sin 50^\circ $$ 8. Solve for $b$: $$ b(\cos 50^\circ - \sin 50^\circ) = 55 (\sin 30^\circ - \cos 30^\circ) $$ Calculate values: $\cos 50^\circ \approx 0.6428$, $\sin 50^\circ \approx 0.7660$, $\sin 30^\circ = 0.5$, $\cos 30^\circ \approx 0.8660$ $$ b(0.6428 - 0.7660) = 55 (0.5 - 0.8660)$$ $$ b(-0.1232) = 55 (-0.3660)$$ $$ b = \frac{55 \times (-0.3660)}{-0.1232} = \frac{-20.13}{-0.1232} \approx 163.3\text{ cm} $$ 9. Now find $d$ using one projection: $$ d = 55 \cos 30^\circ + 163.3 \cos 50^\circ = 55 \times 0.8660 + 163.3 \times 0.6428 = 47.63 + 104.95 = 152.58\text{ cm} $$ **Final answer:** The diagonal length is approximately $152.6$ cm.