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Question
m∠1 = m∠6 =
m∠2 = m∠7 =
m∠3 = m∠8 =
m∠4 = m∠9 =
m∠5 =
Step1: Find the value of \(x\)
Since \(\angle1=(4x + 3)^{\circ}\) and \(\angle9=(9x-47)^{\circ}\), and \(\angle1=\angle9\) (isosceles triangle property).
Step2: Calculate \(m\angle1\) and \(m\angle9\)
Substitute \(x = 10\) into \(\angle1=(4x + 3)^{\circ}\) and \(\angle9=(9x-47)^{\circ}\)
\(m\angle1=(4\times10 + 3)^{\circ}=43^{\circ}\), \(m\angle9=(9\times10-47)^{\circ}=43^{\circ}\)
Step3: Calculate \(m\angle3\) and \(m\angle8\)
Since \(AD = BD\) (assumed from the diagram symmetry), \(\angle2=\angle1 = 17^{\circ}\) (isosceles triangle property).
\(m\angle3=180^{\circ}-(17^{\circ}+43^{\circ})=120^{\circ}\), similarly \(m\angle8 = 120^{\circ}\)
Step4: Calculate \(m\angle4\) and \(m\angle6\)
\(m\angle4=180^{\circ}-120^{\circ}=60^{\circ}\), \(m\angle6 = 60^{\circ}\)
Step5: Calculate \(m\angle5\)
Since \(\angle4=\angle6 = 60^{\circ}\), \(\triangle BDE\) is equilateral. So \(m\angle5=60^{\circ}\)
Step6: Calculate \(m\angle7\)
Since \(EC = EB\) (assumed from the diagram symmetry), \(\angle7=\angle9 = 17^{\circ}\)
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\(m\angle1 = 43^{\circ}\), \(m\angle2 = 17^{\circ}\), \(m\angle3 = 120^{\circ}\), \(m\angle4 = 60^{\circ}\), \(m\angle5 = 60^{\circ}\), \(m\angle6 = 60^{\circ}\), \(m\angle7 = 17^{\circ}\), \(m\angle8 = 120^{\circ}\), \(m\angle9 = 43^{\circ}\)