Question

question
in the diagram below, \\( \overline{ad} \parallel \overline{eg} \\), \\( m\angle bfc = 75 ^ { \circ } \\) and \\( m\angle fbc = 43 ^ { \circ } \\). find \\( m\angle gpc \\).

answer: attempt 1 out of 3
\\( m\angle gpc = \square ^ { \circ } \\)

Explanation:

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

Since \(AD\parallel EG\), \(\angle BCF=\angle GFC\) (alternate - interior angles). In \(\triangle BFC\), we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(\angle GPC = x\).
We know that \(\angle FBC + \angle BFC+\angle BCF=180^{\circ}\). Also, \(\angle BFC=\angle BPC + \angle GPC\) (angle - addition property).

Step2: Substitute the known values

We are given that \(\angle FBC = 43^{\circ}\) and \(\angle BPC = 75^{\circ}\). Let \(\angle GPC=x\). Then \(\angle BFC=75^{\circ}+x\).
Since \(\angle BCF=\angle GFC\) (because \(AD\parallel EG\)), and using the angle - sum property of a triangle \(\angle FBC+\angle BFC+\angle BCF = 180^{\circ}\). But \(\angle BCF=\angle GFC\) (alternate - interior angles for \(AD\parallel EG\)).
Another way:
We use the exterior - angle property. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
For \(\triangle BFC\), \(\angle BPC\) is an exterior angle. \(\angle BPC=\angle FBC+\angle GPC\) (by the exterior - angle theorem, since \(AD\parallel EG\) and we can consider the transversal \(BF\) and \(CF\)).

Step3: Solve for \(x\)

Substitute the values into the equation \(\angle BPC=\angle FBC+\angle GPC\).
\(75^{\circ}=43^{\circ}+x\).
Subtract \(43^{\circ}\) from both sides: \(x = 75^{\circ}-43^{\circ}\).

Answer:

\(32\)