In a circle, the measure of an inscribed angle is half the measure of the intercepted arc. Since angle \( ACB \) (the inscribed angle at vertex \( C \)) subtends arc \( AB \), the measure of angle \( ACB \) is half the measure of the central angle \( AOB \) that subtends the same arc.
In triangle \( ABC \), the angles must sum to \( 180^\circ \). Given that angle \( A \) is \( 27^\circ \), we denote angle \( B \) as \( x \).
Using the fact that the quadrilateral formed by points \( A \), \( B \), \( C \), and the center of the circle has angles summing to \( 360^\circ \), we know that:
\[ \text{Angle } A + \text{Angle } B + \text{Angle } C = 180^\circ \]
Since angle \( C \) also forms an inscribed angle subtended by the same arc as angle \( A \) (which is opposite \( B \)), we know that:
\[ \text{Angle } C = \frac{1}{2} \cdot \text{Angle } AOB = 90^\circ - \frac{1}{2} \cdot x \]
From the triangle sum property:
\[ 27^\circ + x + C = 180^\circ \]
We also know that:
\[ C = 90^\circ - \frac{1}{2}x \]
Substituting the expression for \( C \) into the triangle sum equation gives:
\[ 27^\circ + x + \left(90^\circ - \frac{1}{2}x\right) = 180^\circ \]
Combine and solve for \( x \):
\[ 27^\circ + x + 90^\circ - \frac{1}{2}x = 180^\circ \] \[ 117^\circ + \frac{1}{2}x = 180^\circ \] \[ \frac{1}{2}x = 63^\circ \] \[ x = 126^\circ \]
Therefore, the measure of angle \( B \) is \( 126^\circ \).
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