Question

the angle bisectors of △xyz are \\(\overline{xg}\\), \\(\overline{yg}\\), and \\(\overline{zg}\\). they meet at a single point g. (in other words, g is the incenter of △xyz.) suppose \\(dg = 12\\), \\(yg = 15\\), \\(m \angle dye = 94^\circ\\), and \\(m \angle fzg = 18^\circ\\). find the following measures. note that the figure is not drawn to scale. \\(fg = \square\\) \\(m \angle fze = \square^\circ\\) \\(m \angle fxg = \square^\circ\\)

Explanation:

Step1: Find \( FG \)

The incenter \( G \) is equidistant from all sides of the triangle. So, \( DG = EG = FG \). Given \( DG = 12 \), thus \( FG = 12 \).

Step2: Find \( m\angle FZE \)

Since \( ZG \) is an angle bisector, \( m\angle FZG = m\angle EZG = 18^\circ \). So, \( m\angle FZE = m\angle FZG + m\angle EZG = 18^\circ + 18^\circ = 36^\circ \).

Step3: Find \( m\angle FXG \)

First, find \( m\angle DYG \). Since \( YG \) is an angle bisector and \( \angle DYE = 94^\circ \), the sum of angles in a triangle: the angles at \( D \) and \( E \) are right angles (\( 90^\circ \)). So, in quadrilateral \( DGEY \), \( m\angle DYG = 180^\circ - 94^\circ = 86^\circ \)? Wait, no. Wait, \( \angle DYE \) is the angle at \( Y \) between \( DY \) and \( EY \), which are perpendicular to \( XG \) and \( ZG \)? Wait, no, \( D \) is on \( XY \), \( E \) on \( YZ \), \( F \) on \( XZ \), with \( GD \perp XY \), \( GE \perp YZ \), \( GF \perp XZ \). So \( \angle DYE \) is \( 94^\circ \), so the angle at \( Y \) for the incenter: the sum of angles in triangle \( XYZ \): the incenter angles. Wait, the sum of angles in a triangle is \( 180^\circ \). We know \( \angle FZE = 36^\circ \), \( \angle DYE \) is related to angle at \( Y \). Wait, \( \angle DYE \) is \( 94^\circ \), which is the angle between \( DY \) and \( EY \), which are perpendicular to \( XG \) and \( ZG \)? No, \( GD \perp XY \), \( GE \perp YZ \), so \( \angle G D Y = \angle G E Y = 90^\circ \). So in quadrilateral \( D G E Y \), sum of angles is \( 360^\circ \), so \( m\angle D G E = 360^\circ - 90^\circ - 90^\circ - 94^\circ = 86^\circ \). But maybe easier: the angle at \( Y \) in triangle \( XYZ \): \( \angle XYZ = 2 \times \) angle between \( YG \) and \( YZ \). Wait, no, let's find the angle at \( X \). The sum of angles in triangle \( XYZ \): \( \angle X + \angle Y + \angle Z = 180^\circ \). We have \( \angle Z = 36^\circ \), \( \angle Y \): the angle at \( Y \) is \( 180^\circ - 94^\circ \)? Wait, no, \( \angle DYE \) is \( 94^\circ \), which is the angle between \( DY \) and \( EY \), which are sides of the triangle? Wait, maybe better: the angle at \( Y \): \( \angle XYZ \), and \( YG \) bisects it. The angle between \( GD \) and \( GE \) is \( 94^\circ \), so the angle at \( G \) between \( GD \) and \( GE \) is \( 94^\circ \), so the angle at \( Y \) is \( 180^\circ - 94^\circ = 86^\circ \)? Wait, no, in quadrilateral \( D G E Y \), angles at \( D \) and \( E \) are \( 90^\circ \), so \( \angle D G E = 360 - 90 - 90 - 94 = 86^\circ \), but \( \angle D G E \) is equal to \( 180^\circ - \angle XYZ / 2 \)? No, maybe I messed up. Wait, let's use the sum of angles. We have \( \angle Z = 36^\circ \), let's find \( \angle Y \): \( \angle XYZ \), and \( \angle X \). The sum is \( 180 \). The incenter angles: the angle at \( G \) for \( \angle FXG \): \( \angle FXG \) is half of \( \angle X \). So first, find \( \angle Y \): since \( GD \perp XY \), \( GE \perp YZ \), so \( \angle G D Y = \angle G E Y = 90^\circ \), so \( \angle D G E = 180^\circ - \angle XYZ \) (because in quadrilateral \( D G E Y \), sum of angles is \( 360^\circ \), so \( 90 + 90 + \angle XYZ + \angle D G E = 360 \), so \( \angle D G E = 180 - \angle XYZ \)). But \( \angle DYE = 94^\circ \), which is \( \angle D G E \)? Wait, no, \( \angle DYE \) is \( \angle D Y E \), which is the angle at \( Y \) between \( DY \) and \( EY \), which are segments on \( XY \) and \( YZ \). So \( \angle XYZ = \angle D Y E = 94^\circ \)? Wait, that makes sense. So \( \angle XYZ = 94^\circ \), \( \angle X Z Y = 36^\circ \) (from step 2). So \( \angle X = 180^\circ - 94^\circ - 36^\circ = 50^\circ \). Then \( \angle FXG \) is half of \( \angle X \), so \( m\angle FXG = 50^\circ / 2 = 25^\circ \). Wait, let's check: \( \angle X = 50^\circ \), so angle bisector \( XG \) splits it into two \( 25^\circ \) angles. Yes, that works.

Answer:

\( FG = \boxed{12} \)

\( m\angle FZE = \boxed{36}^\circ \)

\( m\angle FXG = \boxed{25}^\circ \)