Question

finding angle measures complete the statement given that: ( mangle fhe = mangle bhg = mangle ahf = 90^circ ) if ( mangle dhf = 133^circ ), then ( mangle chg = ) °.

Explanation:

Step1: Find \( m\angle DHE \)

Since \( m\angle FHE = 90^\circ \) (given), and \( \angle DHF \) and \( \angle DHE \) are supplementary? Wait, no. Wait, \( \angle DHF \) is composed of \( \angle DHE \) and \( \angle FHE \)? Wait, no, looking at the diagram, \( \angle FHE = 90^\circ \), so \( \angle DHF = \angle DHE + \angle FHE \)? Wait, no, actually, \( \angle DHE \) is a right angle? Wait, no, the given is \( m\angle FHE = 90^\circ \), so \( HE \) is perpendicular to \( HF \). So \( \angle DHE \) is part of \( \angle DHF \). Wait, \( m\angle DHF = 133^\circ \), and \( m\angle FHE = 90^\circ \), so \( m\angle DHE = m\angle DHF - m\angle FHE = 133^\circ - 90^\circ = 43^\circ \)? Wait, no, maybe I got the angles wrong. Wait, actually, \( \angle CHG \) and \( \angle DHE \) are vertical angles? Wait, no, let's re-examine.

Wait, the given angles: \( m\angle FHE = m\angle BHG = m\angle AHF = 90^\circ \). So \( HF \perp HE \), \( BH \perp HG \), \( AH \perp HF \). So \( HE \) is horizontal? Wait, \( AHE \) is a straight line (since \( A \) and \( E \) are on a horizontal line), and \( HF \) is vertical (since \( \angle AHF = 90^\circ \)), so \( HF \perp HE \), \( HF \perp AH \). Similarly, \( BHG = 90^\circ \), so \( BH \perp HG \).

Now, \( \angle DHF = 133^\circ \), and \( \angle DHE \) is adjacent to \( \angle FHE \) (which is \( 90^\circ \)). Wait, \( \angle DHF = \angle DHE + \angle FHE \)? No, \( \angle FHE = 90^\circ \), so \( \angle DHE = \angle DHF - \angle FHE = 133^\circ - 90^\circ = 43^\circ \). But \( \angle DHE \) and \( \angle CHG \): are they equal? Wait, \( \angle DHE \) and \( \angle CHG \): let's see the vertical angles or complementary. Wait, \( \angle CHG \): since \( \angle BHG = 90^\circ \), and \( \angle BHC + \angle CHG = 90^\circ \)? No, maybe \( \angle DHE \) and \( \angle CHG \) are equal because of vertical angles or some other relation. Wait, actually, \( \angle DHE \) and \( \angle CHG \): let's look at the diagram. The lines: \( CD \) and \( AB \) or \( GF \)? Wait, maybe \( \angle DHE \) and \( \angle CHG \) are vertical angles? Wait, no, \( \angle DHE \) and \( \angle CHG \): let's see, \( H \) is the vertex. \( \angle DHE \) has sides \( HD \) and \( HE \), \( \angle CHG \) has sides \( HC \) and \( HG \). Wait, maybe \( \angle DHE \) and \( \angle CHG \) are equal because \( \angle DHF = 133^\circ \), and \( \angle CHF = 90^\circ \) (since \( \angle AHF = 90^\circ \) and \( AHC \) is a straight line? No, \( AHE \) is horizontal, \( HF \) is vertical, so \( \angle CHF = 90^\circ \) (since \( HC \) is vertical? Wait, \( C \) is on the vertical line with \( F \), so \( HC \) is vertical, \( HF \) is vertical, so \( HC \) and \( HF \) are the same line? Wait, no, the diagram shows \( C \) and \( F \) on a vertical line, so \( HC \) and \( HF \) are colinear, downward from \( H \) to \( F \), upward to \( C \). So \( \angle CHF = 180^\circ \), but \( \angle AHF = 90^\circ \), so \( \angle AHC = 90^\circ \), meaning \( HC \) is vertical, \( AH \) is horizontal, so \( HC \perp AH \).

Now, \( \angle DHF = 133^\circ \), and \( \angle CHD \) is adjacent to \( \angle CHG \)? Wait, maybe another approach: \( \angle DHF = 133^\circ \), and \( \angle CHF = 90^\circ \) (since \( HC \perp HF \)? Wait, no, \( HC \) and \( HF \) are the same line, so \( \angle CHF = 180^\circ \). Wait, I think I made a mistake. Let's start over.

Given \( m\angle FHE = 90^\circ \), so \( HE \perp HF \). \( \angle DHF = 133^\circ \), so \( \angle DHE = \angle DHF - \angle FHE = 133^\circ - 90^\circ = 43^\circ \). Now, \( \angle DHE \) and \( \angle CHG \): are they vertical angles? Wait, \( \angle DHE \) and \( \angle CHG \): let's see, \( \angle DHE \) is formed by \( HD \) and \( HE \), \( \angle CHG \) is formed by \( HC \) and \( HG \). Since \( HC \perp AH \) (because \( \angle AHF = 90^\circ \) and \( HC \) is vertical) and \( HE \perp AH \) (since \( AHE \) is horizontal), so \( HC \) and \( HE \) are both vertical and horizontal? No, \( HC \) is vertical, \( HE \) is horizontal, so \( HC \perp HE \), so \( \angle CHE = 90^\circ \). Wait, \( \angle DHE = 43^\circ \), so \( \angle DHC = \angle CHE - \angle DHE = 90^\circ - 43^\circ = 47^\circ \)? No, that doesn't make sense. Wait, maybe \( \angle CHG \) is equal to \( \angle DHE \) because of vertical angles. Wait, \( \angle DHE \) and \( \angle CHG \): if \( HD \) and \( HC \) are on a line, and \( HE \) and \( HG \) are on a line, then they would be vertical angles. Wait, the diagram shows \( CD \) as a line, and \( AB \) or \( GF \) as another line? Maybe \( CD \) is a straight line, so \( \angle CHD = 180^\circ \), but no. Wait, the key is that \( \angle DHF = 133^\circ \), and \( \angle FHG = 90^\circ \) (since \( \angle BHG = 90^\circ \) and \( BHG \) and \( FHG \) are related? Wait, \( \angle BHG = 90^\circ \), so \( BH \perp HG \), and \( \angle AHF = 90^\circ \), so \( AH \perp HF \). So \( AH \) and \( HE \) are horizontal, \( HF \) and \( HC \) are vertical. So \( \angle DHF = 133^\circ \), so the angle between \( HD \) and \( HF \) is \( 133^\circ \), and since \( HF \) is vertical, the angle between \( HD \) and the vertical (HC) is \( 133^\circ - 90^\circ = 43^\circ \). Then \( \angle CHG \) is equal to that angle because \( HG \) is horizontal? Wait, no, \( HG \) is... Wait, maybe \( \angle CHG = 180^\circ - 133^\circ = 47^\circ \)? No, that's not right. Wait, let's use the fact that \( \angle DHF + \angle FHG = 180^\circ \)? No, \( \angle DHF = 133^\circ \), and \( \angle FHG = 90^\circ \) (since \( \angle BHG = 90^\circ \) and \( FHG \) is the same as \( BHG \)? No, \( \angle BHG = 90^\circ \), so \( BH \perp HG \), and \( \angle AHF = 90^\circ \), so \( AH \perp HF \). So \( AH \) and \( HE \) are horizontal, \( HF \) and \( HC \) are vertical. So \( \angle DHF = 133^\circ \), so the angle between \( HD \) and \( HF \) (vertical) is \( 133^\circ \), so the angle between \( HD \) and \( HE \) (horizontal) is \( 133^\circ - 90^\circ = 43^\circ \). Then \( \angle CHG \) is equal to that angle because \( HG \) is horizontal and \( HC \) is vertical, so \( \angle CHG = 43^\circ \). Yes, that makes sense. So \( m\angle CHG = 43^\circ \).

Step2: Confirm the angle relationship

Since \( \angle DHF = 133^\circ \) and \( \angle FHE = 90^\circ \), subtracting gives \( \angle DHE = 133^\circ - 90^\circ = 43^\circ \). Then, because \( \angle DHE \) and \( \angle CHG \) are vertical angles (or due to the perpendicular lines and angle relationships), \( m\angle CHG = 43^\circ \).

Answer:

43