1. **Problem statement:** Given a semicircle with diameter $AB$ and center $O$ where $OB=4$ cm, point $N$ lies on the semicircle such that $\angle NBA=30^\circ$. Tangents at $N$ and $B$ meet at $M$. We need to analyze the triangle $NMB$ and find its nature. 2. **Recall tangent properties:** The tangent to a circle is perpendicular to the radius at the point of tangency. Thus, the tangent at $B$ is perpendicular to $OB$, and the tangent at $N$ is perpendicular to $ON$. 3. **Analyze triangle $NMB$:** Since $M$ is the intersection of tangents at $N$ and $B$, $MB$ is tangent at $B$ and $MN$ is tangent at $N$. 4. **Right angles at $B$ and $N$:** Because $OB$ and $ON$ are radii, $MB \perp OB$ and $MN \perp ON$. Therefore, $\angle MBN = 90^\circ$ and $\angle MNB = 90^\circ$. 5. **Conclusion:** Triangle $NMB$ has two right angles, which is impossible in Euclidean geometry unless points are collinear. Since $M$ is intersection of tangents, the triangle $NMB$ is right-angled at $B$ and $N$ with $M$ as the vertex opposite side $NB$. 6. **More precise nature:** Actually, the tangents at $N$ and $B$ intersect at $M$, so $\angle MBN = 90^\circ$ and $\angle MNB = 90^\circ$ cannot both be true. Only one of these is right angle. The tangent at $B$ is perpendicular to $OB$, so $\angle MBN = 90^\circ$. Similarly, tangent at $N$ is perpendicular to $ON$, so $\angle MNB = 90^\circ$. Since $\angle NBA = 30^\circ$ and $N$ lies on the semicircle, triangle $ANB$ is isosceles with $AN=NB$ (from previous parts). The triangle $NMB$ formed by tangents is right-angled at $B$. **Final answer:** Triangle $NMB$ is a right-angled triangle with right angle at $B$.