# Thread: Supplementary & Complementary Angles

1. ## Supplementary & Complementary Angles

http://i28.tinypic.com/2d0hzzk.jpg

I'm having a bit of trouble with this one guys...

I know Supplementary Angles are Angles which measure up to 180 degrees. I know that ZY+WX = 175. I'm not sure where to get the extra 5 degrees from?

As for the vertical angles, I know the lines ZX, and YK form a vertical Angle. I'm just not sure how to express it.

2. zy+xy=180 (angles on a line =180)
120(zy) + xy=180
xy=60

now the other line's angles are xy, wx and kw which will all equal 180
you know that xy=60, wx=55, kw=?
which all add up to = 180 (angles on a line =180)
60+55+kw=180
180-115=kw
65=kw

check*
55(wx) +65(kw) = 120(zy)(vertically opposite angles are =)
120=120

Hope that helped?

3. Originally Posted by Sarlowbat
http://i28.tinypic.com/2d0hzzk.jpg

I'm having a bit of trouble with this one guys...

I know Supplementary Angles are Angles which measure up to 180 degrees. I know that ZY+WX = 175. I'm not sure where to get the extra 5 degrees from?

As for the vertical angles, I know the lines ZX, and YK form a vertical Angle. I'm just not sure how to express it.

Hi Sarlowbat,

I'm quite baffled at your notation when representing angles formed by intersecting lines.

According to your diagram, this is what I see.

$\displaystyle m\angle WJX=55^{\circ}$

$\displaystyle m\angle ZJY = 120^{\circ}$

If $\displaystyle \overline{ZX} \ \ and \ \ \overline{KT}$ intersect at J, then

$\displaystyle m\angle XJY=180-120 = 60^{\circ}$

$\displaystyle m\angle KJZ = 60^{\circ}$ because $\displaystyle \angle XJY$ and $\displaystyle \angle KJZ$ are vertical angles.

$\displaystyle \angle ZJY$ and $\displaystyle \angle KJX$ are vertical angles as well, so $\displaystyle m\angle KJX=120^{\circ}$

$\displaystyle m\angle WJK=120-55=65^{\circ}$