Question

prom the prom committee is decorating the venue for prom and wants to hang lights above the diagonals of the rectangular room. if $dh = 44.5$ feet, $ef = 39$ feet, and $mangle ghf = 128^circ$, find $mangle hef$.

Explanation:

Step1: Identify rectangle diagonal property

In a rectangle, diagonals bisect each other, so $HE = DH = 44.5$ feet.

Step2: Identify vertical angle

$\angle GHF$ and $\angle EHD$ are vertical angles, so $m\angle EHD = 128^\circ$.

Step3: Find base angles of $\triangle HEF$

First, focus on $\triangle HEF$: $HE = HF = 44.5$ feet, so it is isosceles. First, find $\angle HEF$'s supplementary adjacent angle in the triangle formed by the diagonal.
First, calculate $\angle HEF$'s base angle:
In $\triangle EHD$, $HE = HD$, so base angles are equal:
$m\angle HED = \frac{180^\circ - 128^\circ}{2} = 26^\circ$
Now, $\angle DEF = 90^\circ$ (rectangle corner), so $m\angle HEF = 90^\circ - 26^\circ = 64^\circ$.

Answer:

$64^\circ$