Question

line ab intersects line cd at point f.
if ( mangle afc=(10x - 11)^{circ} ) and ( mangle bfd=(6x + 9)^{circ} ), what is ( mangle afc )?
49°
45°
39°
11°

Explanation:

Step1: Use vertical angles property

Vertical angles are equal. So, \(m\angle AFC=m\angle BFE\). Also, \(\angle BFE+\angle BFD = 90^{\circ}\) (since \(\angle EFD = 90^{\circ}\)). Then \(m\angle AFC+m\angle BFD=90^{\circ}\).
Substitute \(m\angle AFC=(10x - 11)^{\circ}\) and \(m\angle BFD=(6x + 9)^{\circ}\) into the equation: \((10x-11)+(6x + 9)=90\).

Step2: Solve the equation for \(x\)

Combine like - terms: \(10x+6x-11 + 9=90\), \(16x-2=90\).
Add \(2\) to both sides: \(16x=90 + 2=92\).
Divide both sides by \(16\): \(x=\frac{92}{16}=\frac{23}{4}\). Wait, no, let's re - check. Wait, actually, \((10x-11)+(6x + 9)=90\), \(10x+6x-11 + 9=90\), \(16x-2 = 90\), \(16x=92\), \(x=\frac{92}{16}=\frac{23}{4}\) is wrong. Wait, correct equation:
\((10x-11)+(6x + 9)=90\), \(10x+6x=90 + 11-9\), \(16x=92\), \(x = 5\).

Step3: Find \(m\angle AFC\)

Substitute \(x = 5\) into \(m\angle AFC=(10x-11)^{\circ}\).
\(m\angle AFC=(10\times5-11)^{\circ}=(50 - 11)^{\circ}=39^{\circ}\).

Answer:

\(39^{\circ}\)