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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 5.6, Problem 18E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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Exercises 18 and 19 are based on a theorem ( not stated) that is the converse of Theorem 5.6.3. See the figure on above.

Given: N P = 4 , M N = 8 , P Q = 3 , and M Q = 6 ; m ∠ P = 63 ∘ and m ∠ M = 27 ∘
Find: m ∠ P N Q
(HINT: N P M N = P Q M Q . )
Chapter 5.6, Problem 18E, Exercises 18 and 19 are based on a theorem not stated that is the converse of Theorem 5.6.3. See

To determine

To find:

m∠PNQ.

Explanation

Given:

NQ→ bisects ∠MNP, NP=4,MN=8,PQ=3,MQ=6,m∠P=63∘,m∠M=27∘.

Theorem used:

Angle bisector theorem:

If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

Calculation:

Given: NQ→ bisects ∠MNP.

Then by angle-bisector theorem, the ray NQ→ divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

⇒NPMN=PQMQ

In a triangle the sum of the angles is 180∘.

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Chapter 5.6,
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Chapter 5.6,
Problem 19E
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