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Lady bird [3.3K]
2 years ago
10

Pls help if you actually can

Mathematics
1 answer:
Setler79 [48]2 years ago
5 0

The triangles that are similar would be ΔGCB and ΔPEB  due to Angle, Angle, Angle similarity theorem.

<h3>How to identify similar triangles?</h3>

From the image attached, we see that we are given the Parallelogram GRPC. Thus;

A.  The triangles that are similar would be ΔGCB and ΔPEB  due to Angle, Angle, Angle similarity theorem.

B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;

∠CGB ≅ ∠PEB  (Alternate Interior Angles)

∠BPE ≅ ∠BCG  (Alternate Interior Angles)

∠GBC ≅ ∠EBP  (Vertical Angles)

C.  To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;

225/325 = PE/375  

PE = 260 ft

BE/425 = 225/325  

BE = 294 ft

Read more about Similar Triangles at; brainly.com/question/14285697

#SPJ1

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Answer:

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Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.

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180° - 147° = 33°

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