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gtnhenbr [62]
3 years ago
15

Can anyone help me with this? I've been stuck on this question for some time-

Mathematics
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

Angle E is 43 degrees.

Step-by-step explanation:

So we know that AB is parallel to CD.

We also know that Angle CHF is 119 degrees and that Angle B is 18 degrees.

First, let's find Angle DHF. It is the supplementary angle to Angle CHF. Thus, they must total 180:

\angle DHF+\angle CHF=180

We already know that Angle CFH is 119. Substitute:

\angle DHF+119=180

Subtract 119 from both sides:

\angle DHF=61

Therefore, Angle DHF is 61 degrees.

Since Angle DHF is 61 degrees, then angle BGH <em>must</em> also be 61 degrees. This is because they are corresponding angles, and corresponding angles have the same measure.

Also, Angle BGE is the supplementary angle to Angle BGH. In other words:

\angle BGE+\angle BGH=180

Find Angle BGE. Substitute 61 for BGH:

\angle BGE+61=180

Subtract 61:

\angle BGE=119

So, angle BGE is 119 degrees.

Recall that a triangle's interior angles sum to 180. This means that Angle B plus Angle BGE plus Angle E must equal 180. Thus:

\angle B+\angle BGE+\angle E=180

Substitute 18 for B and 119 for BGE. Thus:

18+119+\angle E=180

Add:

137+\angle E=180

Subtract:"

\angle E=43\textdegree

And we're done!

Remarks:

I just realized that you can just use the fact that alternate exterior angles have the same measures to figure out that BGE is equivalent to CHF. Regardless, we will arrive at the same answer :)

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