1. The problem asks to find the measure of angle $B$ in a triangle where the angles at vertices $A$, $B$, and the third vertex are given as $3x$, $2x$, and an algebraic expression respectively. 2. We know the sum of the interior angles of any triangle is $180^\circ$. 3. Assume the third angle is $x^\circ$, then the equation for the sum of angles is: $$3x + 2x + x = 180$$ 4. Simplify the left side: $$6x = 180$$ 5. Solve for $x$: $$x = \frac{180}{6} = 30$$ 6. The angle at vertex $B$ is $2x$, so: $$2 \times 30 = 60^\circ$$ 7. Looking at the given options (18°, 36°, 72°, 54°), $60^\circ$ is not one of them. Let's reconsider the assumption about the third angle. If the third angle corresponds to the unlabeled vertex and equals $4x$ (commonly angle sums in integer multiples), then $$3x + 2x + 4x = 9x = 180$$ $$x = 20$$ Angle $B$ would be: $$2x = 2 \times 20 = 40^\circ$$ Still not in options. Check options mathematically: - If $\angle B = 18^\circ = 2x$, then $x=9$, angles are $3x=27$, $2x=18$, and third angle $=180-27-18=135$ - If $\angle B=36^\circ$, then $x=18$, angles are $3x=54$, $2x=36$, third angle $=90$ - If $\angle B=54^\circ$, then $x=27$, angles are $3x=81$, $2x=54$, third angle $=45$ - If $\angle B=72^\circ$, then $x=36$, angles are $3x=108$, $2x=72$, third angle $=0$ Only $36^\circ$ option gives a valid triangle. Final answer: $\boxed{36^\circ}$