In circle J below, what is the value of x?
central angles have the same measure as the arc they cut...so since the angle LJK cuts arc LK, and mLK = 156, mLJK = 156 as well so \[156=3x-6\] solve for x
x=54
looks good to me :)
These seem so simple once you spell it all out to me lol.
lol...geometry's not too bad once you gather a few rules
I have a couple more like these. Care to help?
sure...i've got time
ok...this one is fun :) maybe i'm just crazy... :P
It's okay I find math fun sometimes too. X)
couple things to add to your toolbox... we already know that central angles measure the same as the arc they cut...but these aren't central angles if it's *not* a central angle (but with vertex on the circle), then the angle is *half* the measure of the arc...still not what we have... so what *DO* we have?!?
i'm glad you asked...
we know that mXLY = 148...so we can find mYLZ (what is it?) also, we know that angle YLZ = angle XLW (why?)
Ehh I'm not sure.. This unit is really confusing to me.
well angle YLZ =angle XLW because they're vertical angles right?
exactly... and we can find mYLZ since it is 180 - mXLY since they form a straight line...
but actually, i just noticed there's an easier way
So YLZ and XLW= 32
yes...but it turns out we don't need those measures...we need another tool... notice that angle XLY = angle WLZ as they are also vertical anthese angles are not central angles and they don't have vertices on the circle...but the measure of these angles is the average of the measure of the arcs they cut...
probably a little confusing, so let me explain better...
mXLY = mWLZ = 148 (mXY + mWZ)/2 = 148
are you fine with that, or do you need it clearer?
Maybe a little clearer
So the angles marked A have the same measure because they are vertical angles, yes? |dw:1400955994433:dw|
Mhmm
sorry...screwed up the numbers
let's say P is 36 and Q is 200, their average\[\frac{36+200}{2}=\frac{236}{2}=118\] is the measure of angle A
Okay
thsis is true for any interior anglese...and if you think about it, it leads to the fact about central angles and angles whose vertex is on the circle too...
more comfortable now?
Yeah
good...so how does this help? we have an expression for arcs XY and WZ...we can take the average of them...this will equal the angle XLY whch we know is 148.
\[\frac{(7x-9) + 3(2x+15)}{2}=148\] solve for x
20?
with confidence :)
20!
Haha I have one more.
bring it on!
i know it's one of two answers...need to dust off a couple cobwebs...1 sec :)
Mmkay
got it... so here we have an external angle. the idea is very similar, but whereas, we add and divide by two for internal angles, we *subtract* and divide by 2 for external angles
meaning \[\frac{mAD-mBC}{2}=mP\]
30!
so remember: (1/2)(sum of arcs) = internal angle (1/2)(difference of arcs) = external angle
i like the confidence :) correct!
Thank you so much!!
you are quite welcome...anytime
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