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This is on a practice assessment, not for a grade therefore I need help solving this because I completely forgot what to do

This Is On A Practice Assessment Not For A Grade Therefore I Need Help Solving This Because I Completely Forgot What To Do class=

Sagot :

Since line DB is perpendicular to line AC, then arc AB = arc BC

Thus, 14x - 3 = 7x + 18

Collecting like terms,

14x - 7x = 18 + 3

7x = 21

Dividing both sides by 7, we have

[tex]\begin{gathered} x=\frac{21}{7} \\ x=3 \end{gathered}[/tex][tex]\begin{gathered} \angle AC=\angle AOB+\angle BOC \\ \angle AC=14x-3+7x+18 \\ \end{gathered}[/tex]

substituting x into the above equation,

[tex]\begin{gathered} \angle AC=14(3)-3+7(3)+18 \\ \angle AC=42-3+21+18 \\ \angle AC=78^0 \end{gathered}[/tex]

Hence, angle the arc AC makes at the centre is 78 degrees.

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