Really odd problem in geometry?

[Frogeyedpeas]

I'm not sure why but I can't seem to figure this problem out at all...

You have a slice of a banked circular track... Find the angle theta. See attachment,

My first guess was to use the pythagorean theorem + the fact that the two lengths are both equal to angle/360* 2 * pi * r where r is different for both lengths but is related by pythagorean theorem however that has been inconclusive

The ratio of the lengths has not been helpful either.

 

[dodo]

It seems to me that there is missing data; either the smaller radius or the pie's angle.

 

[the-ever-kid]

It seems to me that there is missing data; either the smaller radius or the pie's angle.

i think dodo's right there is something missing...either angle of curvature or radius

 

[NascentOxygen]

You don't think that diagram is meant to represent one quadrant of the circular track?

 

[Frogeyedpeas]

nah my teacher has told me that's all I need... idk I'm just going to ask him to walk me through it becuz i really don't see it

 

[the-ever-kid]

let me tell you why we actually need more data...

first let's collect all the data given...

[itex]Arc_b = 97[/itex]
[itex]Arc_s = 74[/itex]
[itex]Height = 12[/itex]

So to get the angle [itex]\theta[/itex] we need to use anyone of the trig ratios:

[itex]tan\theta = \frac{height}{base}[/itex] , [itex]sin\theta = \frac{height}{hypotenuse}[/itex] or [itex]cos\theta = \frac{base}{hypotenuse}[/itex]

[flatten out the image.]

the closest we have is two arcs that could probably give us the base:

let us assume that the angular part of the whole circular curve that we took is [itex]\alpha[/itex]

as [itex]\frac{Arc}{radius} = \alpha[/itex]

[itex]\therefore \frac{Arc_b}{r_1} = \frac{Arc_s}{r_2}[/itex] ,[itex] i.e. r_1 = 1.31081 r_2[/itex]

now [itex]r_1 - r_2 = .3108 r_2[/itex]

[itex]\therefore tan\theta = \frac{12}{.3108\dot r_2}[/itex]

so you need either [itex]\alpha~ or~ r_2~ or~ r_1[/itex]

 

[NascentOxygen]

You don't think that diagram is meant to represent one quadrant of the circular track?

Assuming this figure is OP's attempt at sketching ¼ of the velodrome,
outer radius of track=circumference ÷ (2π) = 4*97÷ (2π) = 61.752
inner radius of track = 4*74 ÷ (2π) = 47.109

So base is (61.752 - 47.109) = ...

 

[agent_509]

I can't seem to think of how, but it seems to me that that is all you need to solve this problem. Thinking a little less rigorously if you had this hypothetical section of track, you couldn't change the base or hypotenuse without affecting one of the known values. Perhaps I'm wrong though.

 

[emailanmol]

As the-Ever-Kid has pointed out, you cannot solve the sum without one more piece of data.(i.e either one radius or angle )

Be assured, it can't be found with the currenly available piece of data :-)


Nascent Oxygen has assumed its one-fourth.(although it doesn't look like one fourth to me :P :D )
You can solve the problem and see you will get the answer.

You can then solve for if it was 2-5th and still get another answer whuch suggests it really depends on amount of alpha ( or radius which you show by a similar reasoning)and that a variable is missing.

 

[agent_509]

ok I see where I made an error in my reasoning now. Yeah I can see why this couldn't be solved without more data now.

 

[the-ever-kid]

hey there people i just pmed frogeyedpeas and yes his teacher told him that it was to be assumed as a quarter part of the track

 

[2slowtogofast]

I know this is a bit old but why would this not work

Both arc 97 and 74 share the same angle so

97 = r1*β

74 = r2*β

2 equation 3 unknws we need another eqn

Looking at the right triangle

sinθ = 12 / HYP

HYP can = r1-r2 so

sinθ = 12/r1-r2

Need to relate β and θ

Here is how i tried that

[PLAIN]http://i1230.photobucket.com/albums/ee500/tosloww/diagram.gif[/PLAIN]

Uggh sorry for the terrible diagram laptop mouse is messed but anyway

from there you could say

β = (180 - 2θ)

Now we have 3 eqn and 3 unknws

97 = r1 (180 - 2θ)

74 = r2 (180 - 2θ)

sinθ = 12/r1-r2

Am I way off base with this I am kinda tired but I still wanted to give it a go

 

[the-ever-kid]

Am I way off base with this I am kinda tired but I still wanted to give it a go

Yes you are! lolz ...see you are going and relating angles in entirely diferent planes you can't do that...coz' only angle between straight lies can be equated not angle between perpendicular lines ...the angle β that you considered is on the zx plane while the angle θ is in the zy plane..