find ( mangle bfc ).( mangle bfc = square^circ )

Answer

Explanation:

Step1: Identify right angle and angle relation

Line (AF) is perpendicular to (GC), so (\angle AFC = 90^\circ). Also, (\angle BFA) and (\angle EFD) are vertical angles (or we can use the fact that (\angle EFD = 28^\circ) and (\angle BFC) relates to the right angle). Wait, actually, since (AF\perp GC) (right angle at (F) between (AF) and (GC)), and the angle between (FD) and (EF) is (28^\circ), the angle between (BF) and (FC) can be found by noting that (\angle AFD = 90^\circ) (since (AF\perp FD) as (AF) and (GC) are horizontal - vertical, so (AF) is vertical, (GC) is horizontal, (FD) is vertical down). Wait, maybe better: the angle between (BF) and (AF) should be equal to the angle between (EF) and (FD) because they are vertical angles? Wait, no. Let's see: (AF) is vertical up, (FD) is vertical down, so (AFD) is a straight line (180 degrees), but (AF\perp GC), so (\angle AFC = 90^\circ) (right angle). The angle between (EF) and (FD) is (28^\circ), so the angle between (BF) and (AF) is also (28^\circ) (vertical angles: (\angle BFA) and (\angle EFD) are vertical angles, so they are equal). Then, since (\angle AFC = 90^\circ), which is (\angle BFA + \angle BFC), so (\angle BFC = 90^\circ - \angle BFA). Since (\angle BFA = 28^\circ) (because (\angle EFD = 28^\circ) and they are vertical angles), then (\angle BFC = 90 - 28 = 62^\circ)? Wait, no, wait: maybe I got the angles reversed. Let's look at the diagram again: (F) is the intersection point. (GC) is horizontal (left (G), right (C)), (AF) is vertical up, (FD) is vertical down, (EB) is a line through (F), with (E) below, (B) above. The angle between (EF) and (FD) is (28^\circ) (so (\angle EFD = 28^\circ)). Then (\angle BFA) is equal to (\angle EFD) because they are vertical angles (opposite angles when two lines intersect). So (\angle BFA = 28^\circ). Now, (\angle AFC) is a right angle (90 degrees) because (AF) is perpendicular to (GC) (the little square at (F) between (AF) and (GC) indicates a right angle). So (\angle AFC = \angle BFA + \angle BFC = 90^\circ). Therefore, (\angle BFC = 90^\circ - \angle BFA = 90 - 28 = 62^\circ)? Wait, no, wait: if (\angle EFD = 28^\circ), then (\angle BFA = 28^\circ) (vertical angles). Then (\angle BFC) is adjacent to (\angle BFA) and together they make (\angle AFC = 90^\circ) (right angle). So (\angle BFC = 90 - 28 = 62)? Wait, no, maybe I have the angle direction wrong. Wait, maybe the angle between (BF) and (FC) is (90 - 28 = 62)? Wait, no, let's think again. The right angle is between (AF) and (GC) (so (\angle AFC = 90^\circ)). The line (EB) crosses (AF) and (FD). The angle between (EF) and (FD) is (28^\circ), so the angle between (BF) and (AF) is also (28^\circ) (vertical angles). Then, since (\angle AFC = 90^\circ = \angle BFA + \angle BFC), then (\angle BFC = 90 - 28 = 62^\circ). Wait, but maybe I mixed up the angles. Alternatively, the angle between (BF) and (FD) is (90^\circ) (since (AF) is vertical, (FD) is vertical, so (BF) makes an angle with (FD) of (90 - 28 = 62)? No, maybe the correct approach is: since (AF\perp GC), (\angle AFC = 90^\circ). The angle (\angle BFA) is equal to (\angle EFD = 28^\circ) (vertical angles). Therefore, (\angle BFC = \angle AFC - \angle BFA = 90^\circ - 28^\circ = 62^\circ).

Step2: Calculate the angle

So (m\angle BFC = 90 - 28 = 62) degrees.

Answer:

(62)