1. **State the problem:** We need to find the measure of angle $B$ (denoted as $m\angle B$) in triangle $ABC$ where the angles are given as follows: - Angle $A = 60^\circ$ - Angle $B = 6x^\circ$ - External angle at $C = (10x + 20)^\circ$ 2. **Recall the external angle theorem:** The external angle at vertex $C$ is equal to the sum of the two opposite internal angles $A$ and $B$. So, $$ 10x + 20 = 60 + 6x $$ 3. **Solve for $x$:** $$ 10x + 20 = 60 + 6x $$ Subtract $6x$ from both sides: $$ 10x - \cancel{6x} + 20 = 60 + \cancel{6x} $$ $$ 4x + 20 = 60 $$ Subtract 20 from both sides: $$ 4x + \cancel{20} - \cancel{20} = 60 - 20 $$ $$ 4x = 40 $$ Divide both sides by 4: $$ \frac{4x}{\cancel{4}} = \frac{40}{\cancel{4}} $$ $$ x = 10 $$ 4. **Find $m\angle B$:** $$ m\angle B = 6x = 6 \times 10 = 60^\circ $$ **Final answer:** $m\angle B = 60^\circ$