Properties Of Triangles

The governing relation is the law of sines, which states that in any triangle the ratio of each side to the sine of its opposite angle is constant, written $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. This proportionality lets an unknown side be recovered as soon as one matching side-angle pair and the second angle are known. Rearranging the relevant proportion for $b$ gives $b=\frac{a\sin B}{\sin A}$. Substituting the standard values $\sin A=\sin 30^{\circ }=\frac{1}{2}$ and $\sin B=\sin 45^{\circ }=\frac{\sqrt{2}}{2}$ along with $a=10$, we obtain $b=\frac{10\cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}}=\frac{5\sqrt{2}}{\frac{1}{2}}=10\sqrt{2}$. Option $5\sqrt{2}$ forgets to divide by $\sin 30^{\circ }$, halving the result. Option $20$ incorrectly replaces $\sin 45^{\circ }$ with $1$. Option $10\sqrt{3}$ uses $\sin 60^{\circ }$ in place of $\sin 45^{\circ }$. This matches the canonical sine-rule application in JEE Advanced triangle problems. As a final plausibility check, the larger angle lies opposite the larger side, and since $B=45^{\circ }>A=30^{\circ }$, the side $b$ must exceed $a$; indeed $10\sqrt{2}\approx 14.1>10$, confirming the answer is geometrically consistent.