Question

∠bfd is a straight angle and ∠afb≅∠efd. if m∠efd=(7x + 8)°, m∠dfc=(16x - 21)°, and m∠efc=(13x + 27)°, find m∠afb.

Explanation:

Step1: Recall angle - addition property

Since \(\angle BFD\) is a straight - angle and \(\angle AFB\cong\angle EFD\), and \(\angle DFC+\angle EFC = 180^{\circ}\) (linear pair), and \(\angle AFB+\angle DFC+\angle EFC = 180^{\circ}\) (because \(\angle BFD = 180^{\circ}\)). Also, given \(\angle AFB=\angle EFD=(7x + 8)^{\circ}\), \(\angle DFC=(16x-21)^{\circ}\), and \(\angle EFC=(13x + 27)^{\circ}\).

Step2: Set up the equation

\((7x + 8)+(16x-21)+(13x + 27)=180\).

Step3: Combine like terms

First, combine the \(x\) - terms: \(7x+16x + 13x=36x\), and combine the constant terms: \(8-21 + 27=14\). So the equation becomes \(36x+14 = 180\).

Step4: Solve for \(x\)

Subtract 14 from both sides of the equation: \(36x=180 - 14=166\). Then \(x=\frac{166}{36}=\frac{83}{18}\).

Step5: Find \(m\angle AFB\)

Since \(m\angle AFB=(7x + 8)^{\circ}\), substitute \(x = \frac{83}{18}\) into the expression for \(m\angle AFB\):
\(m\angle AFB=7\times\frac{83}{18}+8=\frac{581}{18}+\frac{144}{18}=\frac{581 + 144}{18}=\frac{725}{18}\approx40.28^{\circ}\).

Answer:

\(m\angle AFB=\frac{725}{18}\text{ degrees}\)