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Homework Help: Swedish Math problem i got. Seems to be easy. but not quite ?

  1. Mar 30, 2014 #1
    Swedish Math problem i got. Seems to be easy. but not quite....?

    This is a problem i chose because i thought it would be interesting as well as easy, however, i should have known better than to choose the last question in my maths book as my "essay" question. I have done it and i cant see how my method is wrong in any way. The answer i get is not realistic hence, something must be wrong. I will not show my calculations as it would take too long.

    You are in a movie theatre. The screen is 8 m high and 2 meters from the ground as well as 3 meters from the first row of seats. All the seats are on an incline att an angle of 22 degrees. When you sit on a chair your eyes are 1 meter from the incline. Where on the incline are you to sit so that you will have the best viewing angle?
    This is how it looks like and how i have thought a bit!

    The text translates to "what value of "h" will give the best possible viewing angle? i.e how far upp the incline must you walk?"
    Answer i got was 2.56 meters. Any help would be appreciated.
  2. jcsd
  3. Mar 30, 2014 #2
    Would you not want your eye to be level with the centre of the screen?
  4. Mar 30, 2014 #3
    That seems reasonable. However i would need to prove that having the eye level at the centre gives the highest angle, something which i have no clue how to do. Should i use a^2=b^2+c^2 - 2bc(cos(A)) ?
  5. Mar 30, 2014 #4


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    That formula could be useful in some way but first you have to find out what you want to calculate.

    Did you draw a sketch?

    In the real world, probably. Here, not.
  6. Mar 30, 2014 #5
    I want to calculate the highest possible value for Z, which you can see here. I differentiated a bit at the end when i got a function of h. I probably forgot to mention that it is necessary to have a function f(h)=Z
  7. Mar 30, 2014 #6
    Mfb, please remind me never go and see a movie with you, lest I end up with a stiff neck, lol!
  8. Mar 30, 2014 #7
    If we are allowed stiff necks then.....
    Call the distance below the centre of the screen a, split z into z1 and z2 above and below the horizontal axis, derive expressions for z1 and z2. Then use.....
    Tan z = tan z1 + tan z2/(1-tan z1tan z2)
    Then I would Differentiate to find the maximum of tan z.
    Is that what you did?
  9. Mar 30, 2014 #8
    This is not what i did. Interesting, i will try it out and get back to you. Thanks!
  10. Mar 30, 2014 #9
    Ok i am lost. I have no idea what you did there.
  11. Mar 30, 2014 #10


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    I doubt this is what is wanted. They probably want the position where the screen subtends the greatest angle at the eye.
  12. Mar 30, 2014 #11

    Exactly! Any ideas?
  13. Mar 30, 2014 #12


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    What do you know about angles subtended by chords of a circle?
  14. Mar 30, 2014 #13
    I know the basics. I might post my method tomorrow as i just cant figure out what is wrong with it, but i am interested to know how you are thinking!
  15. Mar 30, 2014 #14


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    Draw a circle passing through the top and bottom of the screen and through a position on the seating ramp. What position maximises the angle?
  16. Mar 31, 2014 #15
    I drew it all very detailed and the problem is that the circle only touches a part at the bottom of the extended hypotenuse!
  17. Mar 31, 2014 #16
    I got that the eye level needs to be 63 cm lower than the bottom of the screen. We must be looking for a seat really near the front, less than 4m away from the screen. Stiff neck and nausea!
  18. Mar 31, 2014 #17
  19. Mar 31, 2014 #18
    Now i understand! Very nice, thanks ! however, i dont see how this works if the triangle in red is representing the angle of sight. in my head none of the equations would work out. Would appreciate an explanation! THANKS!
  20. Mar 31, 2014 #19
    Ah, yes that occurred to me too, but I think the tan(a+b) formula must still hold for negative angles. The blue lines are what I drew initially setting up the problem and the red line are the solution I got by maximising tan (a+b).
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