   Chapter 5.5, Problem 29E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem   To determine

To find:

NQ, in the MNQ, which is an equianglular with NR=6 and NR bisects MNQ, QR bisects. MQN

Explanation

Approach:

For a right triangle, for which the measure of the interior angles 30°, 60°, and 90°; if ‘a’ is the length of measure of the shorter leg; opposite to the angle 30°, then the length of the other two sides is given by

Length of the longer leg (opposite to 60°) =a3

Length of the hypotenuse (opposite to 90°)=2a.

In general

Length of the longer leg =3× (Length of the shorter leg)

Length of the hypotenuse =2× (Length of the shorter leg)

Calculation:

Given,

A equiangular triangle MNQ with NR bisects MNQ, QR bisects MNQ.

NR=6.

MNQ, is an equiangular triangle.

Thus,

mM=60°

mN=60°, and

mQ=60°

Now, the angle MNQ and MQN are bisected by the rays NR and QR respectively.

mRNQ=12·mMNQ

=12(60°)

mRNQ=30°

Also,

c12·mMQN

=12(60°)

mRQN=30°

Now, consider the right triangle RNQ of the above diagram.

We have

mN=mQ=30°

Thus, by the theorem that the measure of the sides opposite to the congruent angles are congruent, we have

NR¯RQ¯

Thus,

NR=RQ=6 units

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