1. **State the problem:** We need to find the measure of angle $\angle BCD$ in the given triangle and external angle setup. 2. **Identify known angles and variables:** - $\angle A = 53^\circ$ - $\angle B = (6x - 15)^\circ$ - External angle at $C$ is $(11x - 12)^\circ$ 3. **Recall the external angle theorem:** The external angle at vertex $C$ equals the sum of the two opposite interior angles, $\angle A$ and $\angle B$. 4. **Set up the equation:** $$ 11x - 12 = 53 + (6x - 15) $$ 5. **Simplify the right side:** $$ 11x - 12 = 53 + 6x - 15 $$ $$ 11x - 12 = 6x + 38 $$ 6. **Isolate $x$ by subtracting $6x$ from both sides:** $$ \cancel{11x} - 12 = \cancel{6x} + 38 + (11x - 6x) $$ $$ 5x - 12 = 38 $$ 7. **Add 12 to both sides:** $$ 5x - \cancel{12} + 12 = 38 + 12 $$ $$ 5x = 50 $$ 8. **Divide both sides by 5:** $$ \frac{5x}{\cancel{5}} = \frac{50}{\cancel{5}} $$ $$ x = 10 $$ 9. **Find $\angle B$ using $x=10$:** $$ \angle B = 6(10) - 15 = 60 - 15 = 45^\circ $$ 10. **Find $\angle BCD$:** Since $\angle BCD$ is the external angle at $C$, it equals $$ \angle BCD = 11x - 12 = 11(10) - 12 = 110 - 12 = 98^\circ $$ **Final answer:** $$\boxed{98^\circ}$$