Question

in the diagram below, \\( \overline { a d } \parallel \overline { e g } \\), \\( m \angle b f c = 7 1 ^ { \circ } \\) and \\( m \angle g f c = 3 6 ^ { \circ } \\). find \\( m \angle f b c \\).

step
angle
reason
1
\\( m \angle b f c = 7 1 ^ { \circ } \\)
given
2
\\( m \angle g f c = 3 6 ^ { \circ } \\)
given
\\( \mathrm { m } \angle \square = \square ^ { \circ } \\)
select a reason

Explanation:

Step1: Use the property of parallel lines and alternate - interior angles

Since \(\overline{AD}\parallel\overline{EG}\), we know that \(\angle FBC\) and \(\angle BFE\) are related. First, find \(m\angle BFE\). We know that \(\angle BFG=\angle BFC+\angle GFC\). Given \(m\angle BFC = 71^{\circ}\) and \(m\angle GFC=36^{\circ}\), then \(m\angle BFG=71^{\circ}+ 36^{\circ}=107^{\circ}\). But we need to use the triangle - angle sum or the parallel - line property. Another way: By the alternate - interior angles theorem (if two parallel lines are cut by a transversal, alternate - interior angles are equal). Let's consider the transversal \(BF\) cutting \(\overline{AD}\) and \(\overline{EG}\). We know that in \(\triangle BFC\), we can also use the fact that \(\angle FBC+\angle BFC+\angle BCF = 180^{\circ}\), but since \(\overline{AD}\parallel\overline{EG}\), \(\angle BCF=\angle GFC\) (alternate - interior angles).

Step2: Calculate \(m\angle FBC\)

Using the angle - sum property of a triangle (\(\angle FBC+\angle BFC+\angle BCF = 180^{\circ}\)). Since \(\angle BCF = \angle GFC\) (alternate - interior angles, because \(\overline{AD}\parallel\overline{EG}\) and \(CF\) is a transversal), and \(m\angle BFC = 71^{\circ}\), \(m\angle GFC = 36^{\circ}\). Substitute into the formula \(\angle FBC=180^{\circ}-\angle BFC-\angle BCF\). Since \(\angle BCF=\angle GFC\), we have \(\angle FBC=180^{\circ}-71^{\circ}-36^{\circ}\).

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Answer:

\(m\angle FBC = 73^{\circ}\)