Question

what is the measure of angle wzy?
○ 54.5°
○ 71°
○ 125.5°
○ 180°
(diagram: circle with center x, points w, y on the circle, point z outside the circle connected to w and y, with angle at z between w and y labeled 109°)

Explanation:

Step1: Recall the property of tangents

Tangents from a common external point to a circle are equal in length, so triangle \( WZY \) is isosceles with \( ZW = ZY \). The sum of angles in a triangle is \( 180^\circ \).

Step2: Calculate the base angles

Let \( \angle WZY = \angle ZWY = x \) (wait, no, actually, the angle at \( Z \) is what we need? Wait, no, the angle given is at \( Z \)? Wait, no, the diagram: \( Z \) is the external point, \( ZW \) and \( ZY \) are tangents, so \( \angle WZY \) is the angle between the two tangents. Wait, the arc \( WY \) (the minor arc) is related? Wait, no, the angle between two tangents is equal to \( 180^\circ - \) the measure of the central angle subtended by the same arc. Wait, the central angle \( \angle WXY \): since \( ZW \) and \( ZY \) are tangents, \( \angle XWZ = \angle XYZ = 90^\circ \) (tangent is perpendicular to radius). So quadrilateral \( XWZY \) has two right angles at \( W \) and \( Y \), so the sum of angles in a quadrilateral is \( 360^\circ \). So \( \angle WXY + \angle WZY + 90^\circ + 90^\circ = 360^\circ \). Wait, but maybe easier: in triangle \( WZY \), since \( ZW = ZY \) (tangents from \( Z \) to circle \( X \)), triangle \( WZY \) is isosceles. Wait, the angle at \( Z \) is given as \( 109^\circ \)? No, wait the diagram shows \( 109^\circ \) as the angle at \( Z \)? Wait no, the label is \( 109^\circ \) between \( WZ \) and \( YZ \)? Wait no, the diagram: \( X \) is the center, \( W \) and \( Y \) are on the circle, \( Z \) is outside, \( ZW \) and \( ZY \) are tangents, and the arc \( WY \) (the minor arc) has central angle? Wait, maybe I misread. Wait the problem is to find \( \angle WZY \). Wait, no, maybe the angle at \( Z \) is what we need, but the options include \( 71^\circ \), \( 54.5^\circ \), etc. Wait, let's correct: the angle between two tangents from an external point is equal to \( 180^\circ - \) the measure of the central angle over the intercepted arc. Wait, if the intercepted arc \( WY \) has central angle \( \theta \), then the angle between tangents \( \angle WZY = 180^\circ - \theta \). But maybe the triangle \( WZY \) is isosceles, and the angle at \( Z \) is what we need. Wait, no, the sum of angles in a triangle is \( 180^\circ \). If \( ZW = ZY \), then the base angles are equal. Wait, maybe the angle given is the central angle? Wait, no, the diagram shows \( 109^\circ \) as the angle at \( Z \)? No, the label is \( 109^\circ \) between \( WZ \) and \( YZ \)? Wait, no, the user's diagram: \( Z \) is the vertex, \( W \) and \( Y \) are the other two vertices, with \( 109^\circ \) at \( Z \)? No, that can't be. Wait, maybe the central angle \( \angle WXY \) is \( 109^\circ \), and we need to find the angle at \( Z \). Wait, tangents from \( Z \) to circle \( X \), so \( XW \perp ZW \) and \( XY \perp ZY \), so \( \angle XWZ = \angle XYZ = 90^\circ \). Then in quadrilateral \( XWZY \), sum of angles is \( 360^\circ \), so \( \angle WXY + \angle WZY + 90^\circ + 90^\circ = 360^\circ \), so \( \angle WZY = 360^\circ - 90^\circ - 90^\circ - \angle WXY \). If \( \angle WXY = 109^\circ \), then \( \angle WZY = 360 - 90 - 90 - 109 = 71^\circ \). Ah, that makes sense. So step by step:

Step1: Recall quadrilateral angle sum

Sum of interior angles in a quadrilateral is \( 360^\circ \).

Step2: Identify right angles

\( \angle XWZ = 90^\circ \) and \( \angle XYZ = 90^\circ \) (tangent is perpendicular to radius).

Step3: Let central angle \( \angle WXY = 109^\circ \) (from diagram)

Step4: Calculate \( \angle WZY \)

\( \angle WZY = 360^\circ - \angle XWZ - \angle XYZ - \angle WXY \)
\( = 360^\circ - 90^\circ - 90^\circ - 109^\circ \)
\( = 360 - 289 = 71^\circ \)

Answer:

\( 71^\circ \) (corresponding to the option "71°")