Question

1. if m∠boc = 112°, o is the center of the circle bp is tangent, find m∠pbc.
2. find the measure of arc ab.
3. given ab = 9 and tangent to circle d, and b find the length of ce

Explanation:

Problem 9

Step1: Recall tangent-radius property

A tangent to a circle is perpendicular to the radius at the point of tangency, so \( OB \perp BP \), meaning \( \angle OBP = 90^\circ \). Also, \( OB = OC \) (radii of the same circle), so \( \triangle BOC \) is isosceles. The base angles of an isosceles triangle are equal. Let \( \angle OBC=\angle OCB = x \).
Using the angle - sum property of a triangle (\( \angle BOC + \angle OBC+\angle OCB=180^\circ \)) and given \( \angle BOC = 112^\circ \), we have \( 112^\circ+2x = 180^\circ \).
Solving for \( x \): \( 2x=180^\circ - 112^\circ=68^\circ \), so \( x = 34^\circ \), i.e., \( \angle OBC = 34^\circ \).

Step2: Find \( \angle PBC \)

Since \( \angle OBP = 90^\circ \) and \( \angle OBP=\angle OBC+\angle PBC \), we can write \( \angle PBC=\angle OBP-\angle OBC \).
Substituting \( \angle OBP = 90^\circ \) and \( \angle OBC = 34^\circ \), we get \( \angle PBC=90^\circ - 34^\circ = 56^\circ \).

Step1: Recall straight - line angle

Points \( A \), \( O \), and \( C \) are collinear (since \( AC \) is a diameter), so \( \angle AOB+\angle BOC = 180^\circ \) (linear pair of angles).

Step2: Calculate \( m\overset{\frown}{AB} \)

Given \( \angle BOC = 112^\circ \), and the measure of an arc is equal to the measure of its central angle. Let \( m\overset{\frown}{AB}=\angle AOB \).
Using \( \angle AOB+\angle BOC = 180^\circ \), we substitute \( \angle BOC = 112^\circ \): \( \angle AOB=180^\circ - 112^\circ = 68^\circ \). So \( m\overset{\frown}{AB}=68^\circ \).

Answer:

\( 56^\circ \)

Problem 10