Trig Equations and identities

Hey folks I am new here and seeking some math help as it is spring break my teacher is not around, well here is the question.

The London eye is 135 at it's highest point. at It's lowest point it is about 3m off the ground. It takes about 30 minutes to revolve entirely. Across the river, Big ben is about 96m tall. If you were to take a ride on the london eye, for how long were you higher then bigben?.

This is in my Pre calc 12 trig identities and equations assignment, seems a little difficult for the grade level,

Thanks in advance.

Re: Trig Equations and identities

I made a lot of assumptions, so I will list them here

1) The London eye is a circle

2) The London eye moves at a constant rate

3) The ground between the London Eye and Big Ben is considered level enough that other considerations are negligable (such as the curvature of earth)

4) Correct me at any time

I drew a picture; I'll leave it to you to verify the easy calculations. I would say that solving for the unknown angle PMR would be sufficient. Divide the unknown angle into 360 and you would get the proportion of time it takes to rotate along that arc. Finally multiply that proportion by 30 minutes.

http://imageftw.com/uploads/20130322/Wheel.png

Re: Trig Equations and identities

hmm sorry to ask but would you be able to elaborate step by step what you did, i would really appreciate it. I can solve for PMR but I don't understand why or how you got there :(. Sorry for the stupid question.

This is what i got is this for Theta 65 degrees then use this formula to find arc length A= (Theta)(R) (65)(66) arc = 4346.249 Then i times that by two to give me the full arc=8692.49 now how do i interpret this number seconds? or minutes?.

Please correct me if i am wrong.

Re: Trig Equations and identities

The height off the ground is 135. This means that the total diameter of the circle is 132. Dividing 2 gives us the radius, which is 66, so the distance between M (the midpoint) and P (the intersection of the height of B.B) is also 66. The distance from the top of the wheel to point R (which is the same height as B.B) is 135-96=39, so the distance from R to M is 66-39=27.

Now it takes 30 minutes to go a full rotation, in other words, 30 minutes / 360 degrees = 1 minute/12 degrees. Multiply by 65 degrees and you get 5.41 minutes.

EDIT:// whoops, you need to double that time since it goes for the other side of the wheel as well

Re: Trig Equations and identities

I'd model it with a sine function in the form y=a sin b(x-c) +d, where a is the amplitude (half distance between highest and lowest height), the period is 2pi/b (here the period is 30 min), phase shift is c (don't need to consider a phase shift here) and vertical shift is d (height of the centre of the wheel.

Have you learnt that method?

Re: Trig Equations and identities

Quote:

Originally Posted by

**Debsta** I'd model it with a sine function in the form y=a sin b(x-c) +d, where a is the amplitude (half distance between highest and lowest height), the period is 2pi/b (here the period is 30 min), phase shift is c (don't need to consider a phase shift here) and vertical shift is d (height of the centre of the wheel.

Have you learnt that method?

that was in the last chapter yes i have learnt that method and honestly i just don't remember how to get beyond reading the data amplitude period i get all that but get lost from there.

Re: Trig Equations and identities

Ok. So tell mewhat you think the amplitude is, what the period is and what the vertical shift is. Then we'll go from there.