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In rectangle EFGH, EH = 3 and EF = 4. Let M be the midpoint of \(\overline{EF}\), and let X be a point such that MH = MX and \(\angle MHX = 72^\circ\), as shown below. Find \(\angle XGH\), in degrees.

 Jun 4, 2020
 #1
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This is very simple. All you need is a basic trig and the pythagorean theorem.  indecision

 

 Jun 4, 2020
edited by Guest  Jun 4, 2020
 #2
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In rectangle EFGH, EH = 3 and EF = 4. Let M be the midpoint of \(\overline{EF}\),
and let X be a point such that MH = MX and \angle \(MHX = 72^\circ\), as shown below.
Find \(\angle XGH\), in degrees.

\(\begin{array}{|rcll|} \hline 2x &=& 180^\circ-2*72^\circ \quad | \quad : 2\\ x &=& 90^\circ-72^\circ \\ \mathbf{x} &=& \mathbf{18^\circ} \\ \hline \end{array}\)

 

laugh

 Jun 4, 2020

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