KieronN482450 KieronN482450
  • 03-11-2022
  • Mathematics
contestada

In the circle below BD = 172° and ACFind m/BXD.

In the circle below BD 172 and ACFind mBXD class=

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KallinW314827 KallinW314827
  • 03-11-2022

Given:

[tex]\begin{gathered} \hat{BD}=172\degree \\ \hat{AC}=56\degree \end{gathered}[/tex]

To determine the m[tex]m\angle BXD=\frac{1}{2}(\hat{BD}+\hat{AC})[/tex]We plug in what we know:

[tex]\begin{gathered} \begin{equation*} m\angle BXD=\frac{1}{2}(\hat{BD}+\hat{AC}) \end{equation*} \\ m\angle BXD=\frac{1}{2}(172+56) \\ Calculate \\ m\angle BXD=114\degree \end{gathered}[/tex]

Therefore, the answer is:

[tex]\begin{equation*} 114\degree \end{equation*}[/tex]

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