A circle has a diameter AB

and an inscribed angle at vertex C
forming triangle ABC
. The measure of angle A
is 27°. Using what you know about inscribed angles, what is the angle measure of B
?(1 point)
m∠B=

1 answer

In a circle, the measure of an inscribed angle is half the measure of the arc that it subtends. Since \( AB \) is the diameter of the circle, angle \( ACB \) subtends the arc from point \( A \) to point \( B \), which is a semicircle.

According to the inscribed angle theorem, the angle subtended by a semicircle (angle \( ACB \)) is \( 90^\circ \). Therefore, we have:

\[ \angle ACB = 90^\circ \]

Now, in triangle \( ABC \), the angles A, B, and C must sum to \( 180^\circ \):

\[ \angle A + \angle B + \angle C = 180^\circ \]

Substituting the known values:

\[ 27^\circ + \angle B + 90^\circ = 180^\circ \]

Simplifying this gives:

\[ \angle B + 117^\circ = 180^\circ \]

Subtracting \( 117^\circ \) from both sides:

\[ \angle B = 180^\circ - 117^\circ = 63^\circ \]

Thus, the measure of angle \( B \) is:

\[ \boxed{63^\circ} \]