Inscribed Angle help! @e.mccormick
Ok, so, where is the angle?
Line AB is tangent to circle O at B. Find the length of the radius r for AB = 8 and AO = 11.7. Round to the nearest tenth if necessary. The diagram is not to scale.
I know we have to use the pithagorian theorum.
a^2+b^2=c^2 but I dont know which order the numbers go in...
Because AB is tangent, yes, you can use pythagorus. Now, in that C is the Hypotenuse. That help? Or need a little more?
Let me check
Also, the tangent means that the angle at B is the right angle.
C= AO?
Yah.
What are the other two angles called? I can't remember their names. Hypotenus and the other two? Once I know their names I can plug them into the equation. But for some reason I cannot think of them lol.
they just call the other sides the legs.
The angles opposite them have no special names... other than they are acute because it is a right triangle.
Oh. so it doesn't matter which one is a or b?
Exactly.
Is this right?: 8^2+r^2=11.7^2
Then I just simplify?
Well, solve for r.
64+r^2=136.89
I got r=8.5?
Yah, that is rounded to 1 decimal.
Lol yeah, they asked for the nearest tenth.
Yep. So that works.
Would you mind helping me with another problem?
Sure, what is it?
Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale.
Do you see where the right triangle is hiding that exposes the measure of all radii?
I see the right angle, yeah.
Add a line to make the right triangle. That will be the hypotenuse you need to solve for.
Okay, so: 9^2+6^2=x^2?
whoops i mean 12^2 instead of the 6^2
because that line on the top of the circle is broken into half, and since one side is 12, the other side is also 12. Right?
81+144=x^2 225=x^2 Then I square rooted both to get 15=x
Okay, so those ones are getting easier for me. But there's these other ones I have to do and they look difficult.
I am thinking they gave you \(12^2+9^2=?^2\) because visuially it does liik that way. So yah, 15.
I really appriciate you helping me, that last time you helped me I got a 98% on my test.
Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale.
x would equal 40 right? because they are congruent?
hello?
Had a bit of lag there... fun connection at times. Dude, did you read the Code of Conduct for the site? http://openstudy.com/code-of-conduct One of the things they are against is things that "...violates your school's Honor Code..." like getting someone to help you with a test.
No you didn't help me with a test. You helped me do good on the test. Like, you helped me understand how to do a certain problem that was going to be on the test.
OH, I see.. you meant last time... Aaaah... Kk. That makes more sense now,. Sorry bout that.
There have been a rash of people posting test questions recently. Not at all good.
yes, they would be congruent, so 40%.
Lol, no problem. I don't see the point in cheat because it won't help in the long run.
This is just a worksheet that get assigned to BEFORE the actual test. They don't explain it very well for me to understand, so I found this site and its helped so far!
Yah. There was this sign in an engineering department someone posted to the web, "If you get the calculation wrong in real life, people die. So don't cheat and learn the material!"
Lol yeah. I have 4 more problems then Ill be good to go. Can you help me with them still? Or are you busy?
Sure. Fire away.
Awesome. Thank you.
Find the value of x. If necessary, round your answer to the nearest tenth. The figure is not drawn to scale.
Hmmm.... have any other info?
FG = 20, RS = 24, OP = 14
OK, with OP and half of FG, you can get r. That make sense?
I make it into a triangle again?
Yep!
And once you have r, that can solve the other triangle for OQ or x.
I got 17.2 for r
yah
I got 12.3 for x!
Does that upside down T symbol mean intersect?
Perpendicular.
Oh okay.
And inside a circle, things that are perpendicular to a chord and go through the center, well, they bisect the chord.
and make a 90 degree angle
yes, which is why the right triangle laws can take over.
so \(2\sqrt{38}\approx 12.3\) is great there.
Oh okay.
Ready for the next one? It looks different from the ones we've been doing..
Sure.
Use the diagram. Line AB is a diameter, and Line AB is perpendicular to Line CD . The figure is not drawn to scale.
Then it says find mBD for mAC=58 degrees.
Instead of a line above the letters, it's like a curved line
Arch I think.
Yes, the arc.
OK.... a couple ways to take this on, but, the best start is that you look at this, and with the perpendicular lines, you have an inscribed kite, right?
Yeah, a kite.
Do you remember the rules about what is similar in angles in kites, side lengths, etc?
No...
The only thing I think I remember would be that the two triangles on the bottom are congruent and the ones on the top are congruent also.
and the angles are equal when they are diagonally across from each other, right?
Ah, OK... that is the key to this. because AB is a diameter and perpendicular to CD, we know it is a kite where CP=CD. Because of that, AC=AD. Well, this translates to the arcs inscribing the kite as well!
Okay, so AC and AD both are 58 degrees
Yes, which tells you how much about the rest of the circle?
Circle is 360 degrees so I would add 58+58 and get 116 degrees. Then 360-116=244 degrees?
Yes. The other half is 244.... and the large part of the kite also has equal pieces, so each is half that remainging value.
Half of 244?
122?
So, CB=DB?
Yay. that would be it!
Haha, cool! That was pretty easy.
Once you find one side, you can find the others.
Okay, 2 more!
Find the measure of angle BAC. The figure is not drawn to scale.
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