# Chapter 5.5, Problem 29E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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 30Â°, 60Â°, and 90Â°; if â€˜aâ€™ is the length of measure of the shorter leg; opposite to the angle 30Â°, then the length of the other two sides is given by

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

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

In general

Length of the longer leg =3Ã— (Length of the shorter leg)

Length of the hypotenuse =2Ã— (Length of the shorter leg)

Calculation:

Given,

A equiangular triangle MNQ with NRâƒ‘ bisects âˆ MNQ, QRâƒ‘ bisects âˆ MNQ.

NR=6.

âˆ†MNQ, is an equiangular triangle.

Thus,

mâˆ M=60Â°

mâˆ N=60Â°, and

mâˆ Q=60Â°

Now, the angle âˆ MNQ and âˆ MQN are bisected by the rays NRâƒ‘ and QRâƒ‘ respectively.

mâˆ RNQ=12Â·mâˆ MNQ

=12(60Â°)

mâˆ RNQ=30Â°

Also,

c12Â·mâˆ MQN

=12(60Â°)

mâˆ RQN=30Â°

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

We have

mâˆ N=mâˆ Q=30Â°

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

NRÂ¯â‰…RQÂ¯

Thus,

NR=RQ=6 units

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