Trig questions. The answer deadline is up so I cannot submit the answers anymore, but feel free to answer and help out if you can. A) The figure below is a simplified model of a Ferris wheel with diameter 230 feet. The top of the wheel is 240 feet above the ground. If θ is the central angle formed as a rider moves from position P0 to position P1, find the rider's height above the ground h when θ is 120°. Diagram: http://prntscr.com/8kn5ag B) Suppose each edge of the cube shown in the figure is 9.00 inches long. Find the measure of the angle formed by diagonals CF and CH. Round your answer to the nearest tenth of a degree. Diagram: http://prntscr.com/8lbcvd **Variable version: Suppose each edge of the cube shown in the figure is x inches long. Find the measure of the angle formed by diagonals DE and DG. Round your answer to the nearest tenth of a degree. Diagram http://prntscr.com/8lbe3e

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Trig questions. The answer deadline is up so I cannot submit the answers anymore, but feel free to answer and help out if you can. A) The figure below is a simplified model of a Ferris wheel with diameter 230 feet. The top of the wheel is 240 feet above the ground. If θ is the central angle formed as a rider moves from position P0 to position P1, find the rider's height above the ground h when θ is 120°. Diagram: http://prntscr.com/8kn5ag B) Suppose each edge of the cube shown in the figure is 9.00 inches long. Find the measure of the angle formed by diagonals CF and CH. Round your answer to the nearest tenth of a degree. Diagram: http://prntscr.com/8lbcvd **Variable version: Suppose each edge of the cube shown in the figure is x inches long. Find the measure of the angle formed by diagonals DE and DG. Round your answer to the nearest tenth of a degree. Diagram http://prntscr.com/8lbe3e

Mathematics
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Okay give me some minutes to figure out the solution
Alright
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Almost finished
Gotcha
Take your time, I've got other problems to work on in the meantime :)
@Hero I may be able to solve question A. Are you okay with helping me solve the other two questions?
scratch that I got stuck
I thought I'd be able to solve this but I haven't worked with it in a While and It's very late here. I don't know why you wait until now to want to do this We and have done it at any point during the day.
I just woke up an hour ago. Lol
I'm going to post the solution but it will take a bit longer Just be patient please.
Okay
i think B can be answered using vectors...
The deadline for the question submission is up.
if u set it up so that the cube lies in the first octant with C at the origin, then CH = <9,9,0> and CF is <9,9,9>
oh
No, go ahead. I would still like to know the answer.
...except this class has not taught vectors yet. I do know vectors, but the class hasn't taught us vectors yet. At least I don't think they have... But you can proceed. I know vectors so I'll just use vectors :p
oh ok... then never mind. But the next step would be to use "dot product" in case you are curious. its a formula that allows us to find the angle between any two vectors. you will definitely come across it :)
No it's okay go ahead
I forgot how to dot-product actually... it's been two years ^^;
if you have two vectors A= and B= then the dot product of A and B is (a1b1+a2b2 + a3b3) OR (magnitude of A)(magnitude of B)( cosine of angle between them) so, \[81+81+0= \sqrt{9^2 +9^2 +0} \times \sqrt{9^2 +9^2 +9^2}\times \cos(\theta)\]
solve for theta 35.2 degrees
does it match with the answer you got?
Um... I was working on my psych homework lol I might have to go, so if you wanna do this tomorrow?
no prob
Thanks :D
I have the solution. I will post in five minutes Or less.
Hero, take your time. My answer submission form is already expired so I am fine with waiting however line for the solution. :]
Well, the solution is for part A that's all I have time for sorry.
however long* Okay that's fine you can do it later :]
You can do B without vectors quite simply as well, |dw:1444204348370:dw|
|dw:1444204444350:dw| so we can find CH using python, and its simply \[\sqrt{9^2 + 9^2}\] which simplifies to \[\sqrt{162}\]
|dw:1444204536394:dw| again using pythag \[\tanø = \frac{ 9 }{ √162 }\] therefore ø = 35.2 @kittiwitti1
Also, did you understand part A @kittiwitti1?
Sorry, what's up? Just logged on-

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