Question

given defg is a rectangle, find ( mangle dhg ).
( mangle dhg=)

Explanation:

Step1: Identify equal angles in rectangle

In a rectangle, opposite sides are parallel, so \( \angle DEH \) and \( \angle EFG \) are alternate interior angles? Wait, no. Wait, in rectangle DEFG, \( DE \parallel GF \), and \( DE = GF \), \( DG = EF \). Also, the triangles formed by the diagonals: in a rectangle, diagonals are equal and bisect each other, so \( \triangle DEH \) and \( \triangle FGH \) might be isoceles? Wait, actually, the angles \( \angle DEH = (9x + 3)^\circ \) and \( \angle EFG=(14x - 27)^\circ \). Wait, in a rectangle, \( DE \parallel GF \), so \( \angle DEH \) and \( \angle EFG \) are equal? Wait, no, maybe they are alternate interior angles or something. Wait, actually, in a rectangle, the triangles formed by the diagonals: \( \triangle DEG \) and \( \triangle EFG \)? Wait, no, let's think again. In rectangle DEFG, \( DE \parallel GF \), so \( \angle EDH \) and \( \angle FGH \) are equal? Wait, maybe the angles \( (9x + 3)^\circ \) and \( (14x - 27)^\circ \) are equal because \( DE \parallel GF \) and \( EF \) is a transversal, so alternate interior angles are equal. So set \( 9x + 3 = 14x - 27 \).

Step2: Solve for x

\( 9x + 3 = 14x - 27 \)

Subtract \( 9x \) from both sides: \( 3 = 5x - 27 \)

Add 27 to both sides: \( 30 = 5x \)

Divide by 5: \( x = 6 \)

Step3: Find the measure of one of the angles

Now plug \( x = 6 \) into \( 9x + 3 \): \( 9(6) + 3 = 54 + 3 = 57^\circ \)

Or into \( 14x - 27 \): \( 14(6) - 27 = 84 - 27 = 57^\circ \), so both angles are \( 57^\circ \)

Step4: Find \( m\angle DHG \)

In triangle \( DHG \), or maybe using the fact that \( \angle DHG \) is supplementary to the sum of the two angles? Wait, no. Wait, in the rectangle, the diagonals bisect each other, so \( H \) is the midpoint. Wait, actually, \( \angle DHG \) is an exterior angle or related to the two angles. Wait, the two angles we found are \( 57^\circ \) each, and \( \angle DHG \) is equal to the sum of the two non-adjacent angles? Wait, no, actually, in the triangle, if we consider triangle \( DEH \) and \( FGH \), but maybe \( \angle DHG \) is equal to \( 180^\circ - 2 \times 57^\circ \)? Wait, no, wait. Wait, the angles at \( E \) and \( F \) are \( 57^\circ \), so in triangle \( EHF \), or maybe \( \angle DHG \) is vertical to the angle opposite, but actually, let's think about the triangle. Wait, \( \angle DEH = 57^\circ \), \( \angle EDH \): wait, no, maybe \( \angle DHG \) is equal to \( 180^\circ - (180^\circ - 2 \times 57^\circ) \)? No, that's not right. Wait, actually, \( \angle DHG \) is an angle formed by the diagonals, and since the two angles we found are \( 57^\circ \), then \( \angle DHG = 180^\circ - 2 \times (90^\circ - 57^\circ) \)? No, maybe simpler: in a triangle, the sum of angles is \( 180^\circ \), but wait, actually, \( \angle DHG \) is equal to \( 2 \times 57^\circ \)? Wait, no, let's draw this mentally. The rectangle DEFG, with diagonals DF and EG intersecting at H. So \( \angle DEH = 57^\circ \), \( \angle EFG = 57^\circ \). Then, in triangle \( DEH \), \( \angle EDH = 90^\circ - 57^\circ = 33^\circ \), similarly \( \angle FGH = 33^\circ \). Then \( \angle DHG = 57^\circ + 57^\circ = 114^\circ \)? Wait, no, that doesn't make sense. Wait, actually, \( \angle DHG \) is equal to \( 180^\circ - (180^\circ - 2 \times 57^\circ) \)? No, maybe I made a mistake. Wait, no, let's recall that in a rectangle, the diagonals are equal and bisect each other, so triangles \( DEH \), \( EHF \), \( FHG \), \( GHD \) are isoceles? Wait, no, maybe \( \angle DHG \) is supplementary to the angle we found. Wait, no, let's use the fact that \( \angle DHG \) is equal to \( 180^\circ - (57^\circ + 57^\circ) \)? No, that would be \( 66^\circ \), which is wrong. Wait, no, I think I messed up. Wait, the two angles we found are \( 57^\circ \), which are angles at \( E \) and \( F \). Then, in the triangle formed by \( D \), \( H \), \( G \), the angles at \( D \) and \( G \) would be \( 90^\circ - 57^\circ = 33^\circ \) each, so \( \angle DHG = 180^\circ - 33^\circ - 33^\circ = 114^\circ \). Wait, that makes sense. Because the sum of angles in a triangle is \( 180^\circ \), so if two angles are \( 33^\circ \) each, the third is \( 114^\circ \). Alternatively, since \( \angle DEH = 57^\circ \), and \( \angle EDH = 90^\circ - 57^\circ = 33^\circ \), then \( \angle DHG \) is an exterior angle to triangle \( DHE \), so it's equal to \( \angle DEH + \angle EDH = 57^\circ + 57^\circ = 114^\circ \)? Wait, no, exterior angle is equal to the sum of the two non-adjacent interior angles. So in triangle \( DHE \), \( \angle DHG \) is an exterior angle at \( H \), so it's equal to \( \angle EDH + \angle DEH = 33^\circ + 57^\circ = 90^\circ \)? No, that's not right. Wait, I think I made a mistake in the angle relationships. Let's start over.

Wait, DEFG is a rectangle, so \( DE \parallel GF \), and \( EF \) is a transversal. Therefore, \( \angle DEF + \angle EFG = 180^\circ \)? No, that's if they are same-side interior angles. Wait, no, in a rectangle, all angles are \( 90^\circ \), so \( \angle DEF = 90^\circ \), \( \angle EFG = 90^\circ \), but that's not the case here. Wait, no, the angles given are \( (9x + 3)^\circ \) at \( E \) (angle \( \angle DEH \)) and \( (14x - 27)^\circ \) at \( F \) (angle \( \angle EFG \)). Wait, maybe these angles are equal because \( DE \parallel GF \) and \( EF \) is a transversal, so alternate interior angles are equal. So \( 9x + 3 = 14x - 27 \), which we solved to \( x = 6 \), so the angles are \( 57^\circ \) each. Now, in the rectangle, diagonals \( DF \) and \( EG \) intersect at \( H \), so \( H \) is the midpoint, so \( DH = EH = FH = GH \) (since diagonals in a rectangle are equal and bisect each other). Therefore, triangles \( DEH \), \( EHF \), \( FHG \), \( GHD \) are isoceles. Wait, angle \( \angle DEH = 57^\circ \), so in triangle \( DEH \), \( DH = EH \), so it's isoceles with \( \angle EDH = \angle DEH = 57^\circ \)? No, that can't be, because the sum would exceed \( 180^\circ \). Wait, no, \( \angle DEH \) is at \( E \), between \( DE \) and \( EH \), and \( \angle EDH \) is at \( D \), between \( DE \) and \( DH \). Since \( DH = EH \), then \( \angle EDH = \angle DEH = 57^\circ \), so the third angle \( \angle DHE = 180^\circ - 57^\circ - 57^\circ = 66^\circ \). Then \( \angle DHG \) is supplementary to \( \angle DHE \), so \( 180^\circ - 66^\circ = 114^\circ \). Ah, that makes sense. So \( \angle DHE = 66^\circ \), so \( \angle DHG = 180^\circ - 66^\circ = 114^\circ \).

Wait, let's verify:

If \( x = 6 \), then \( 9x + 3 = 57 \), \( 14x - 27 = 57 \). So angles at \( E \) and \( F \) are \( 57^\circ \). In triangle \( DEH \), \( DH = EH \) (diagonals bisect each other), so it's isoceles with base angles \( 57^\circ \) each. Therefore, vertex angle \( \angle DHE = 180 - 2 \times 57 = 66^\circ \). Then \( \angle DHG \) is vertical to \( \angle EHF \)? No, \( \angle DHG \) and \( \angle DHE \) are supplementary because they form a linear pair. So \( \angle DHG = 180 - 66 = 114^\circ \).

Answer:

\( 114 \)