1. State the problem. In triangle ABC we are given $b=4.2\,\text{cm}$, $c=7.5\,\text{cm}$ and $A=48^\circ36'$. We need to find the remaining side $a$ and angles $B$ and $C$. 2. Formula and important rules. Use the Law of Cosines to find the side opposite $A$: $$a^2 = b^2 + c^2 - 2 b c \cos A$$. After finding a side, use the Law of Sines to find angles: $$\frac{\sin B}{b} = \frac{\sin A}{a}$$. Remember to convert angle minutes to decimal degrees: $48^\circ36' = 48.6^\circ$. 3. Compute $a$ with the Law of Cosines. Convert the angle: $A = 48^\circ36' = 48.6^\circ$. Compute $\cos A = \cos 48.6^\circ \approx 0.661312856$. Plug values into the formula: $$a^2 = 4.2^2 + 7.5^2 - 2\cdot4.2\cdot7.5\cdot\cos 48.6^\circ$$. Evaluate step by step: $4.2^2 = 17.64$. $7.5^2 = 56.25$. $2\cdot4.2\cdot7.5 = 63$. $63\cdot 0.661312856 \approx 41.662709928$. So $a^2 \approx 73.89 - 41.662709928 = 32.227290072$. Therefore $a \approx \sqrt{32.227290072} \approx 5.676809\,\text{cm}$. 4. Find angle $B$ using the Law of Sines. Compute $\sin A = \sin 48.6^\circ \approx 0.750110957$. Then $$\sin B = \dfrac{b\sin A}{a} = \dfrac{4.2\cdot 0.750110957}{5.676809} \approx 0.555001$$. So $B \approx \arcsin(0.555001) \approx 33.723^\circ$. In degrees, minutes, seconds $B \approx 33^\circ43'22.8''$. 5. Find angle $C$ from the angle sum. Compute $C = 180^\circ - A - B = 180^\circ - 48.6^\circ - 33.723^\circ \approx 97.677^\circ$. In DMS $C \approx 97^\circ40'37.2''$. 6. Final answers and check. $a \approx 5.676809\,\text{cm}$. $B \approx 33.723^\circ$ (about $33^\circ43'22.8''$). $C \approx 97.677^\circ$ (about $97^\circ40'37.2''$). Check: $A+B+C \approx 48.6^\circ + 33.723^\circ + 97.677^\circ = 180^\circ$, so the triangle closes.