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. Since \( DG \) is the distance from \( G \) to side \( XY \) (as \( DG \perp XY \)) and \( FG \) is the distance from \( G \) to side \( XZ \) (as \( FG \perp XZ \)), these distances are equal. Given \( DG = 12 \), we have \( FG = DG = 12 \).

Step2: Find \( m\angle FZE \)

\( \overline{ZG} \) is an angle bisector of \( \angle FZE \), so \( \angle FZG=\angle EZG \). Given \( m\angle FZG = 18^\circ \), then \( m\angle FZE=2\times m\angle FZG = 2\times18^\circ = 36^\circ \).

Step3: Find \( m\angle FXG \)

First, find \( m\angle XYZ \). Since \( \overline{YG} \) is an angle bisector and \( \angle DYE = 94^\circ \), \( \angle XYZ = 94^\circ \) (wait, actually \( \angle DYE \) is a straight angle? No, \( \angle DYE \) is formed by \( YD \perp XY \) and \( YE \perp YZ \), so \( \angle DYE = 94^\circ \) implies the angle at \( Y \) between the two perpendiculars. Wait, the sum of angles in a quadrilateral \( DYEG \) is \( 360^\circ \), with \( \angle YDG = \angle YEG = 90^\circ \), so \( \angle DYG + \angle DYE = 180^\circ \)? No, better: the angle at \( Y \), \( \angle XYZ \), since \( YG \) is the angle bisector, and \( \angle DYE = 94^\circ \), actually \( \angle DYE \) is supplementary to \( \angle XYZ \)? Wait, no, \( D \) is on \( XY \), \( E \) is on \( YZ \), \( DG \perp XY \), \( EG \perp YZ \), so quadrilateral \( DYEG \) has two right angles at \( D \) and \( E \). So \( \angle DYG + \angle DYE + \angle YEG + \angle YDG = 360^\circ \), so \( \angle DYG + 94^\circ + 90^\circ + 90^\circ = 360^\circ \), so \( \angle DYG = 360^\circ - 94^\circ - 90^\circ - 90^\circ = 86^\circ \). Then \( \angle XYZ = 2\times\angle DYG = 2\times86^\circ = 172^\circ \)? Wait, no, maybe I messed up. Wait, the sum of angles in a triangle is \( 180^\circ \). We know \( \angle FZE = 36^\circ \) (from step 2), let's denote \( \angle XYZ = y \), \( \angle YXZ = x \), \( \angle XZY = 36^\circ \) (wait no, \( \angle FZE \) is \( \angle XZY \), so \( \angle XZY = 36^\circ \)). Then \( x + y + 36^\circ = 180^\circ \). Now, \( \overline{YG} \) is the angle bisector, and \( \angle DYE = 94^\circ \). Wait, maybe \( \angle DYE \) is the angle between the two perpendiculars, so the angle at \( Y \) between the sides \( XY \) and \( YZ \) is \( 180^\circ - 94^\circ = 86^\circ \)? No, this is confusing. Wait, let's use the fact that in triangle \( XYZ \), the sum of angles is \( 180^\circ \). We have \( \angle XZY = 36^\circ \) (from step 2), let's find \( \angle XYZ \). Wait, \( \angle DYE = 94^\circ \), and \( DG \perp XY \), \( EG \perp YZ \), so \( \angle YDG = \angle YEG = 90^\circ \). So in quadrilateral \( DYEG \), \( \angle DYG + \angle DYE + \angle YEG + \angle YDG = 360^\circ \), so \( \angle DYG + 94^\circ + 90^\circ + 90^\circ = 360^\circ \), so \( \angle DYG = 86^\circ \). Since \( YG \) is the angle bisector, \( \angle XYZ = 2\times\angle DYG = 172^\circ \)? Then \( \angle YXZ = 180^\circ - 172^\circ - 36^\circ = -28^\circ \)? That can't be. Wait, I must have misinterpreted \( \angle DYE \). Maybe \( \angle DYE \) is \( \angle XYZ \)? No, the diagram shows \( D \) on \( XY \), \( E \) on \( YZ \), \( G \) inside the triangle. Wait, maybe \( \angle DYE \) is \( 94^\circ \), so the angle at \( Y \) between \( YD \) and \( YE \) is \( 94^\circ \), and since \( YD \perp XY \) and \( YE \perp YZ \), then \( \angle XYD = 90^\circ \) and \( \angle ZYE = 90^\circ \), so \( \angle XYZ = \angle XYD + \angle ZYE - \angle DYE = 90^\circ + 90^\circ - 94^\circ = 86^\circ \). Ah, that makes sense! So \( \angle XYZ = 86^\circ \). Then, since \( YG \) is the angle bisector, but we need \( \angle YXZ \). The sum of angles in triangle \( XYZ \) is \( 180^\circ \), so \( \angle YXZ + \angle XYZ + \angle XZY = 180^\circ \). We know \( \angle XZY = 36^\circ \) (from step 2) and \( \angle XYZ = 86^\circ \), so \( \angle YXZ = 180^\circ - 86^\circ - 36^\circ = 58^\circ \). Then \( \overline{XG} \) is the angle bisector of \( \angle YXZ \), so \( m\angle FXG = \frac{1}{2}\angle YXZ = \frac{1}{2}\times58^\circ = 29^\circ \).

Answer:

\( FG = \boxed{12} \)
\( m\angle FZE = \boxed{36}^\circ \)
\( m\angle FXG = \boxed{29}^\circ \)