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

  1. #1
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    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.

    Question:
    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!
    https://docs.google.com/drawings/d/1...YC-W7fS44/edit

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

    Can you define what is meant by the best viewing angle ?
    Are we simply looking for the maximum value of z and the corresponding position, or something else ?
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Exactly! The greatest value of Z and its corresponding position in terms of the hypotenuse. Tricky.
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by Cyrus1 View Post
    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.

    Question:
    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!
    https://docs.google.com/drawings/d/1...YC-W7fS44/edit

    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.
    Hej,

    I've modified your sketch a little bit and here is what I've calculated:

    1. The straight line representing the place where your head could be has the equation y = x \cdot \tan(22^\circ) - 3 \cdot \tan(22^\circ)  +1

    z = \alpha+\beta

    2. \tan(\alpha)=\frac{10-y}{x} and \tan(\beta) = \frac{y-2}x

    3. From \tan(z)=\tan(\alpha + \beta)=\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha) \cdot \tan(\beta)} I get:

    \tan(z)=\frac{\frac{10-y}{x} + \frac{y-2}x}{1 - \frac{10-y}{x} \cdot \frac{y-2}x}

    4. Since z has it's maximum if \tan(z) is at it's maximum I solved for x:  \frac{d(\tan(z))}{dx} = 0

    I've got x \approx 4.406798787. That means your head is at (4.406798787, 1.568383604)

    5. ... but honestly I would refuse to take this seat
    Attached Thumbnails Attached Thumbnails Swedish Math problem i got. Seems to be easy. but not quite....?-swedkino.png  
    Thanks from johng and mash
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    How did you derive the equation for y? Thanks!
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    My method was to first make all the sides of the smaller triangle in terms of h. So sin22*h=y (0.375h=y) and cos22 *h=x (0.927h=x). The bigger triangle is a distance, lets say "t", longer horisontaly than the smaller triangle. 1/tan22=2.475=t. at the same time the bigger triangle has a longer hypotenuse, the extra distance being "q". q=1/sin22 = 2.67. Hence, the bigger triangle sides in terms of h is : Hypotenuse = h+2.67, base = 0.927h+2.475 and the height = 0.375h + 1. Now we look at the "angle" triangle and use a^2=b^2+c^2-2bc(cos(A)). a=8 and A is the angle we want maximized. We need to find a function for b and c in terms of "h". if you make a right angled triangle for the side b, where b is the hypotenuse you will easily see that you can get a function for b by using pythagoras theorem. c^2= (0.37h+1-2)^2 + (0.927h+3)^2 and for b^2= (0.927h+3)^2 + (9-0.37h)^2 expand and simplify and also find the equations for just c and b and you will be able to have a function of cos(A). when you have a function of A and put it into wolframalpha i do not get what i was expecting. Do you see any error in this method?
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by Cyrus1 View Post
    How did you derive the equation for y? Thanks!
    Hej,

    you know that the slope of a straight line is the tangens of the angle between the line and the x-axis.

    Use the point-slope-equation of a straight line with the fixed point P(3,0) and the slope m = \tan(22^\circ):

    \frac{y-0}{x-3}=m=\tan(22^\circ)~\implies~y=x \tan(22^\circ)-3 \cdot \tan(22^\circ)

    Now the line is translated one unit up. Therefore the equation is finally: \boxed{y=x \tan(22^\circ)-3 \cdot \tan(22^\circ)+1}

    ********************************

    To your last post: Without a sketch where I can see what you labeled with a, b, c, t, q etc. I'm not able to comprehend your calculations. Sorry.
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Hello over there,
    I made a scale drawing of the described movie screen ( 1cm = 1 m).I found that the level of eye would be 6 m above grade for best viewing.Level of floor at this point would be 5 m above grade. h the length of walk up the rising floor is 5/sin 22 deg. (13.35m)
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    I have it sketched correctly now! If you find the error in my method i will be so grateful!!! btw the angle z is the same as A! Z=A https://docs.google.com/drawings/d/1...YC-W7fS44/edit
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    I am very interested in your method!!
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by earboth View Post
    Hej, 3. From \tan(z)=\tan(\alpha + \beta)=\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha) \cdot \tan(\beta)} I get: \tan(z)=\frac{\frac{10-y}{x} + \frac{y-2}x}{1 - \frac{10-y}{x} \cdot \frac{y-2}x} 4. Since z has it's maximum if \tan(z) is at it's maximum I solved for x:  \frac{d(\tan(z))}{dx} = 0 )
    Btw what did you do with the y variable? I am guessing you wrote it in terms of x before differentiating? second question about your method: Can you tell me how you thought regarding the maximum of tan(z) is the same as maximum of z? THANKS!!!! You are awesome
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by Cyrus1 View Post
    Btw what did you do with the y variable? I am guessing you wrote it in terms of x before differentiating? second question about your method: Can you tell me how you thought regarding the maximum of tan(z) is the same as maximum of z? THANKS!!!! You are awesome
    You can leave $y$ in there until the end, if you want. You can differentiate using the Chain Rule. Earboth did not say that the maximum of $\tan(z)$ is the same as the maximum of z. Earboth said that $\tan(z)$ reaches its maximum when $z$ reaches its maximum. This is because $\tan(z)$ is a strictly increasing function of $z$. However, the maximum viewing angle is not the "best" viewing angle, as it requires you to crane your neck to see the whole screen. I believe the "best" viewing angle would be the distance that simultaneously minimizes both $\alpha$ and $\beta$ in Earboth's sketch, as this would require the least craning of one's neck. So, bjhopper's answer seems most appropriate (post #8 above).
    Thanks from earboth
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by bjhopper View Post
    Hello over there,
    I made a scale drawing of the described movie screen ( 1cm = 1 m).I found that the level of eye would be 6 m above grade for best viewing.Level of floor at this point would be 5 m above grade. h the length of walk up the rising floor is 5/sin 22 deg. (13.35m)
    Hi,

    I've attached my scale drawing. According to post #3 I was looking for the maximum of angle z. I've added the position suggested by bjhopper.

    As SlipEternal pointed out the position is not the best where the maximum angle occurs:
    • your eyes must move very rapidly to cover a wide area
    • all images are distorted because of the perspective
    • you only need a barber to get a shave
    Attached Thumbnails Attached Thumbnails Swedish Math problem i got. Seems to be easy. but not quite....?-swedkinokonstr.png  
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    Quote Originally Posted by earboth View Post
    Hej,


    2. \tan(\alpha)=\frac{10-y}{x} and \tan(\beta) = \frac{y-2}x

    3. From \tan(z)=\tan(\alpha + \beta)=\frac{\tan(\alpha) + \tan(\beta)}{1-\tan(\alpha) \cdot \tan(\beta)} I get:

    \tan(z)=\frac{\frac{10-y}{x} + \frac{y-2}x}{1 - \frac{10-y}{x} \cdot \frac{y-2}x}
    Thanks a lot for everything! The last thing i was wondering is how the method holds for when the X line is under 2 meters. I mean how do you split the angle Z in that case? Will you write
    Last edited by Cyrus1; March 31st 2014 at 11:47 PM.
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    Re: Swedish Math problem i got. Seems to be easy. but not quite....?

    I arrived at the same result as Earboth, but using a different variable.

    Let the length up the slope be h, then the 'head position' will have co-ordinates (h\cos(22)+3, h\sin(22)+1).
    That leads to (using Earboth's notation),
    \tan(\alpha)=\frac{9-h\sin(22)}{h\cos(22)+3} \text{   and   }\tan{\beta}=\frac{h\sin(22)-1}{h\cos(22)+3}.

    Differentiating \tan(z) wrt h and equating that to zero gets you the quadratic
    h^{2}\cos(22) + 6h -30\sin(22)=0 which has the solutions, h=1.517 \text{  or  } h=-7.989.

    The first of these 1.517 leads to x=3+1.517\cos(22)=4.407, same as Earboth.
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