Question

Circle
H
H is shown where points
N
,

Y
,

W
N, Y, W and
K
K are on the circle,
m
N
Y
⏠
=
58
°
,

m
R
K
⏠
=
16
°
,

m
K
W
⏠
=
106
°
,
m
NY
=58°, m
RK
=16°, m
KW
=106°, and
R
Y
‾
RY
is a diameter and a perpendicular bisector of
N
W
‾
.
NW
.








Determine
m
∠
K
X
W
.
m∠KXW.



m
∠
K
X
W
=
m∠KXW=
°
°

Answer 1

To determine \( m \angle KXW \), we can use properties of circles and angles formed by chords and diameters.

Given that \( RY \) is a diameter of the circle, and it is also a perpendicular bisector of \( NW \), the following points about angles can be concluded:

1. Since \( RY \) is a diameter, any angle inscribed in a semicircle is a right angle (90°).
2. Since \( RY \) is the perpendicular bisector of \( NW \), it divides \( NW \) into two equal segments at point \( X \) (the midpoint of \( NW \)). Thus, \( \angle KXN \) and \( \angle KXW \) are equal.

Now let's find \( m \angle KXW \):

1. Since \( m \angle KYN \) is an inscribed angle that intercepts arc \( KN \), we know that \( m \angle KYN \) is half the measure of the intercepted arc:
\[
m \angle KYN = \frac{m N Y}{2} = \frac{58°}{2} = 29°
\]

2. Next, we find \( m \angle KXN \) which should be equal to the angle formed by chords \( KW \) and \( KN \). According to the inscribed angle theorem:
\[
m \angle KXN = \frac{m K W + m K Y}{2}
\]
where:
\[
m KW = 106° \quad \text{and} \quad m KY = 58°
\]
since \( K, N, W \) are on the circle, \( m \angle KYR = 90° \).

3. Now,
\[
m \angle KXN = \frac{106° + 58°}{2} = \frac{164°}{2} = 82°
\]

4. Since \( KX \) is also part of this configuration, due to the symmetry and properties of the circle, \( m \angle KXW \) can also be calculated as follows:
\[
m \angle KXW = 90° - m \angle KYN = 90° - 29° = 61°
\]

Therefore:
\[
m \angle KXW = 82° \quad \text{(due to properties of angles in circles and symmetry)}
\]

Therefore, the final answer is:
\[
m \angle KXW = 82°
\]