Given: Line PB tangent Line PV, Line PU Secants If m Arch VU = 70° and m Arch ST = 30°, then m Angle 2 = a) 20 b) 35 c) 50
The answer I got wrong was 35
average of the two arcs
So, add them together and divide by two?
yes
Okay sorry it took so long
I was Driving off a stalker
I don't think you wanna' know what he said to me either
:(
I'm not worried or anything. Just annoyed
@phi
Report them if it happens again
I will
Don't worry
Oh and btw, there is another question that applies to the same picture
@phi
if it asks for angle 1, that is 1/2 the difference of the two arcs i.e. (big - small)/2
If m Angle 1 = 30° and m Arch ST = 20°, then m Arch VU = a) 10 b) 40 c) 80
\[ \frac{VU- ST}{2} = Angle 1\]
fill in what you know, then use algebra to "solve for VU" can you do that ?
Hmmm. Let me think for a sec
in the formula, replace ST with 20 and angle 1 with 30
\[ \frac{VU- 20}{2} = 30 \] I would start by multiplying both sides by 2
Yeah, I'm writing it down
Oh and thanks. That helps
I was thinking about doing that, but then I started to second guess myself
So VU-40= 60?
VU=100?
you do this \[2\cdot \frac{(VU- 20)}{2} = 30 \cdot 2\] you can think of the left side as multiplying fraction times fraction (top times top and bottom times bottom) \[ \frac{2}{1} \cdot \frac{(VU- 20)}{2} = 30 \cdot 2\] \[\frac{2(VU- 20)}{2} = 30 \cdot 2\] or \[ \frac{2}{2} \frac{(VU- 20)}{1} = 30 \cdot 2\]
in other words you should get VU -20 = 60
Okay. So you don't multiply 20 by 2?
I'm guessing that's where I messed up
remember to how to add or subtract fractions: \[ \frac{VU}{2} - \frac{20}{2} = \frac{VU-20}{2} \] you can do it the other way round: \[ \frac{VU-20}{2}= \frac{VU}{2} - \frac{20}{2} \] and simplify 20/2 to 10 \[ \frac{VU}{2} - 10 = 30 \] now add 10 to both sides \[ \frac{VU}{2} - 10 +10 = 30+10 \\ \frac{VU}{2} = 40 \]
So VU= 80
yes.
Alright I understand it now
I guess you have to get the idea that \[ \frac{complicated\ stuff}{2} \cdot 2 = complicated\ stuff \]
Hehehe. Lol!
wow
I'm gonna' be quoting that for awhile
Thanks!
yw
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