1.8 assignment 9 of 12 finding angle measures complete the statement given that: ( mangle fhe = mangle bhg = mangle ahf = 90^circ ) if ( mangle bhf = 115^circ ), then ( mangle 3 = ) (\boxed{})°.

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# Explanation:
## Step1: Recall that $ m\angle BHG = 90^\circ $ (given).
We know $ m\angle BHF = 115^\circ $, and $ \angle BHF = \angle BHG + \angle GHF $, but actually, we can also look at the relationship between $ \angle BHF $ and the right angle $ \angle FHE = 90^\circ $? Wait, no, better: $ \angle BHF $ and $ \angle 3 $: since $ \angle BHG = 90^\circ $, and $ \angle BHF = 115^\circ $, then $ \angle 3 = \angle BHF - \angle BHG $. Wait, no, wait: $ \angle BHG $ is 90 degrees, so $ \angle BHF = \angle BHG + \angle GHF $? No, maybe another way. Wait, $ \angle FHE = 90^\circ $, so $ HE $ is perpendicular to $ HF $. Wait, the given is $ m\angle FHE = m\angle BHG = m\angle AHF = 90^\circ $. So $ \angle BHG = 90^\circ $, which is a right angle. So $ \angle BHF = 115^\circ $, so the angle adjacent to $ \angle BHG $ in $ \angle BHF $ would be $ \angle BHF - \angle BHG $? Wait, no, $ \angle BHG $ is 90 degrees, so $ \angle 3 $ is $ \angle BHF - 90^\circ $, because $ \angle BHF = \angle BHG + \angle 3 $? Wait, no, let's visualize the diagram. $ H $ is the vertex, with $ BH $, $ CH $, $ DH $, $ EH $, $ FH $, $ GH $, $ AH $. $ \angle BHG = 90^\circ $, so $ BH $ and $ GH $ are perpendicular? Wait, no, $ \angle BHG = 90^\circ $, so $ BH $ and $ HG $ form a right angle. But $ \angle FHE = 90^\circ $, so $ EH $ and $ FH $ are perpendicular. Also, $ \angle AHF = 90^\circ $, so $ AH $ and $ FH $ are perpendicular. So $ AH $ and $ EH $ are a straight line (since $ AHE $ is a horizontal line, as per the diagram: $ A $---$ H $---$ E $ is a straight line). $ FH $ is vertical (since $ \angle AHF = 90^\circ $ and $ \angle FHE = 90^\circ $), so $ FH $ is perpendicular to $ AE $. $ BHG = 90^\circ $, so $ BH $ and $ HG $ are perpendicular, but $ HG $ is along $ FH $? Wait, maybe $ \angle BHF = 115^\circ $, and $ \angle FHE = 90^\circ $, but no, $ \angle FHE $ is 90, so $ EH $ is horizontal, $ FH $ is vertical. Then $ \angle BHF $ is the angle between $ BH $ and $ FH $, which is 115 degrees. Since $ FH $ is vertical (90 degrees from $ EH $), then the angle between $ BH $ and $ EH $ would be $ 115^\circ - 90^\circ = 25^\circ $? Wait, no, maybe $ \angle 3 $ is equal to $ \angle BHF - 90^\circ $, because $ \angle BHG = 90^\circ $, so $ \angle 3 = 115^\circ - 90^\circ = 25^\circ $? Wait, no, let's do it step by step.

Given $ m\angle BHG = 90^\circ $ (given) and $ m\angle BHF = 115^\circ $. We need to find $ m\angle 3 $.

Notice that $ \angle BHF = \angle BHG + \angle GHF $, but no, actually, $ \angle BHG $ is 90 degrees, and $ \angle 3 $ is part of $ \angle BHF $. Wait, maybe $ \angle BHF $ and $ \angle 3 $: since $ \angle BHG = 90^\circ $, then $ \angle 3 = \angle BHF - \angle BHG $. Wait, $ \angle BHF = 115^\circ $, $ \angle BHG = 90^\circ $, so $ \angle 3 = 115^\circ - 90^\circ = 25^\circ $? Wait, no, maybe the other way. Wait, $ \angle FHE = 90^\circ $, so $ EH $ is horizontal, $ FH $ is vertical. $ \angle BHF = 115^\circ $, so the angle between $ BH $ and $ FH $ is 115 degrees. Since $ FH $ is vertical (90 degrees from $ EH $), then the angle between $ BH $ and $ EH $ is $ 115^\circ - 90^\circ = 25^\circ $, but $ \angle 3 $ is equal to that angle? Wait, maybe $ \angle 3 $ is complementary or supplementary. Wait, let's think again.

Wait, $ \angle BHG = 90^\circ $, so $ BH $ is perpendicular to $ HG $. But $ HG $ is along $ FH $ (since $ \angle AHF = 90^\circ $, so $ FH $ is vertical, $ AH $ is horizontal, so $ HG $ is along $ FH $ downward). So $ BH $ and $ FH $ form $ \angle BHF = 115^\circ $, and $ BH $ and $ CH $ (which is vertical upward) form $ \angle BHG = 90^\circ $. Wait, maybe $ \angle 3 $ is $ \angle BHF - 90^\circ $, because $ \angle BHG = 90^\circ $, so $ \angle 3 = 115^\circ - 90^\circ = 25^\circ $. Wait, that makes sense. So:

## Step1: Identify the relationship between angles.
We know $ m\angle BHG = 90^\circ $ (given) and $ m\angle BHF = 115^\circ $. The angle $ \angle 3 $ is equal to $ m\angle BHF - m\angle BHG $ because $ \angle BHF $ is composed of $ \angle BHG $ and $ \angle 3 $ (or vice versa, depending on the diagram).

## Step2: Calculate $ m\angle 3 $.
Substitute the given values: $ m\angle 3 = m\angle BHF - m\angle BHG = 115^\circ - 90^\circ = 25^\circ $.

# Answer:
$ 25 $

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Explanation:

Step1: Recall that \( m\angle BHG = 90^\circ \) (given).

Answer: