Question

lp 6 day 9
task 18
reason quantitatively a wall panel needs to be
cut to fill a space in transition to a staircase as shown
in the figure. if $m\angle dfe = 60$, find $m\angle afd$ and $m\angle fdc$,
the angles at which the panel should be cut.
g.mg.1, smp 2

Explanation:

Step1: Find \( m\angle AFD \)

Angles \( \angle AFD \) and \( \angle DFE \) are supplementary (they form a linear pair, so their sum is \( 180^\circ \)). Given \( m\angle DFE = 60^\circ \), we use the formula \( m\angle AFD + m\angle DFE = 180^\circ \).
\[ m\angle AFD = 180^\circ - 60^\circ = 120^\circ \]

Step2: Find \( m\angle FDC \)

Looking at the figure, \( AF \) and \( ED \) are both vertical (perpendicular to the horizontal line), so \( AF \parallel ED \), and \( FD \) is a transversal. Also, \( \angle DFE = 60^\circ \), and \( \angle FDC \) and \( \angle DFE \) are same - side interior angles? Wait, actually, since \( AE \) and \( ED \) are perpendicular, and \( AF \) is parallel to \( ED \), the figure \( AFDE \) is a rectangle (because \( \angle A = \angle E = 90^\circ \) and \( AF \parallel ED \), \( AE \parallel FD \)? Wait, no, more simply, \( \angle FDC \) and \( \angle DFE \) are supplementary? Wait, no. Wait, the sum of interior angles on the same side of a transversal for parallel lines is \( 180^\circ \), but actually, since \( \angle DFE = 60^\circ \) and \( \angle FDC \) is adjacent to a right - angled side? Wait, no, let's think again. The lines \( AB \) and \( DC \) are parallel (since \( \angle A=\angle B = \angle C=90^\circ \)), and \( FD \) is a transversal. Also, \( \angle AFD = 120^\circ \), and \( \angle AFD \) and \( \angle FDC \) are same - side interior angles? Wait, no, actually, in the figure, \( AFDE \) has \( \angle A=\angle E = 90^\circ \), so \( AF \parallel ED \), and \( FD \) is a transversal. Then \( \angle DFE \) and \( \angle FDC \) are supplementary? Wait, no, \( \angle FDC + \angle DFE=180^\circ - 90^\circ\)? No, wait, let's use the fact that \( \angle FDC \) and \( \angle DFE \) are related to the right angles. Wait, actually, since \( \angle DFE = 60^\circ \), and \( \angle FDC \) and \( \angle DFE \) are supplementary? No, wait, the correct way: since \( AF \) is vertical (perpendicular to the horizontal), and \( DC \) is also horizontal? Wait, no, \( AB \) and \( DC \) are horizontal, \( AF \) and \( BC \) are vertical. So \( AF \parallel BC \), and \( FD \) is a transversal. The angle \( \angle AFD = 120^\circ \), and \( \angle FDC \) and \( \angle AFD \) are same - side interior angles? Wait, no, actually, \( \angle FDC \) and \( \angle DFE \) are supplementary? Wait, no, let's calculate \( \angle FDC \). Since \( \angle DFE = 60^\circ \), and \( \angle FDC \) and \( \angle DFE \) are such that \( \angle FDC=180^\circ - 60^\circ - 90^\circ\)? No, that's wrong. Wait, the correct approach: \( \angle AFD \) and \( \angle DFE \) are supplementary (linear pair), so \( \angle AFD = 120^\circ \). For \( \angle FDC \), since \( AF \parallel ED \) (both vertical), and \( FD \) is a transversal, \( \angle FDC + \angle DFE=180^\circ - 90^\circ\)? No, wait, \( \angle FDC \) and \( \angle DFE \) are same - side interior angles for parallel lines \( AF \) and \( ED \), so \( \angle FDC + \angle DFE = 180^\circ \)? No, \( \angle DFE = 60^\circ \), so \( \angle FDC=180^\circ - 60^\circ=120^\circ \)? No, that can't be. Wait, no, I made a mistake. Let's look at the figure again. The figure has \( \angle A = 90^\circ \), \( \angle B = 90^\circ \), \( \angle C = 90^\circ \), \( \angle E = 90^\circ \). \( AF \) is vertical, \( ED \) is vertical, so \( AF\parallel ED \). Then \( \angle DFE = 60^\circ \), and \( \angle FDC \) and \( \angle DFE \) are supplementary? No, \( \angle FDC \) and \( \angle DFE \) are actually equal? Wait, no. Wait, let's use the property of parallel lines. Since \( AF\parallel ED \), and \( FD \) is a transversal, \( \angle AFD + \angle FDE = 180^\circ \), but \( \angle FDE = 90^\circ \)? No, I'm getting confused. Wait, the correct way: \( \angle AFD \) and \( \angle DFE \) are linear pair, so \( \angle AFD=180 - 60 = 120^\circ \). For \( \angle FDC \), since \( AB\parallel DC \) (because \( \angle A=\angle B=\angle C = 90^\circ \)), and \( FD \) is a transversal, \( \angle AFD + \angle FDC=180^\circ \)? No, that would mean \( \angle FDC = 60^\circ \). Ah, yes! Because \( AB\parallel DC \), and \( FD \) is a transversal, so \( \angle AFD \) and \( \angle FDC \) are same - side interior angles, and their sum is \( 180^\circ \)? Wait, no, if \( \angle AFD = 120^\circ \), then \( \angle FDC=180 - 120 = 60^\circ \). Wait, that makes sense. Because \( AB\parallel DC \), and \( FD \) is a transversal, so \( \angle AFD \) and \( \angle FDC \) are same - side interior angles, so \( \angle AFD+\angle FDC = 180^\circ \). We know \( \angle AFD = 120^\circ \), so \( \angle FDC=180 - 120 = 60^\circ \). Or, since \( \angle DFE = 60^\circ \), and \( \angle FDC \) and \( \angle DFE \) are alternate interior angles? Wait, no, let's start over.

1. For \( \angle AFD \):
- \( \angle AFD \) and \( \angle DFE \) form a linear pair (they lie on a straight line \( AF - FE \)). The sum of angles in a linear pair is \( 180^\circ \).
- Given \( m\angle DFE = 60^\circ \), so \( m\angle AFD=180^\circ - 60^\circ = 120^\circ \).

2. For \( \angle FDC \):
- Lines \( AB \) and \( DC \) are parallel (since \( \angle A=\angle B=\angle C = 90^\circ \), so \( AB\perp BC \) and \( DC\perp BC \), hence \( AB\parallel DC \)).
- \( FD \) is a transversal cutting the parallel lines \( AB \) and \( DC \).
- \( \angle AFD \) and \( \angle FDC \) are same - side interior angles. The sum of same - side interior angles for parallel lines is \( 180^\circ \).
- We know \( m\angle AFD = 120^\circ \), so \( m\angle FDC=180^\circ - 120^\circ = 60^\circ \).

Answer:

\( m\angle AFD = 120^\circ \), \( m\angle FDC = 60^\circ \)