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Solving a Triangle

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    You are trying to find the ideal distance that visitors should stand from the wall to view an art work. The art is 7ft high and is hung so that its bottom is 9ft from the floor. A viewer with eye level at 6 ft stands b ft from the wall. Express θ, the vertical angle subtended by the eye, as a function of b.

    2. Relevant equations
    no idea

    3. The attempt at a solution
    Code (Text):
    so, my main problem is that i don't understand the question :blushing:!!!
    what is the vertical angle subtended by the eye???
    Now, I drew the triangle and found that the vertical side is 10ft, the horizontal side is b ft and thats all i know! How am I supposed to find anything knowing only 1 side???

    Attached Files:

  2. jcsd
  3. Jan 15, 2009 #2
    Use the Cosine Law, I have a picture of what to do, anyhow you should be arriving at the equation:
  4. Jan 15, 2009 #3


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    Just using right triangles should do it. First, of course, draw a picture. You should have a vertical line (the wall) with points marked at 9 feet above the floor and 9+ 7= 13 feet above the floor, marking the top and bottom of the picture. Then draw a person b feet from the wall and at height 6 feet draw a horizontal line from the person's eye to the wall and lines from the person's eye to the top and bottom of the picture.

    You should see two right triangles; the first, the triangle formed by the lines from the person's eye to the wall and from the person's eye to the bottom of the picture, has horizontal base b feet and opposite let 3 feet long. If you let [itex]\theta_1[/itex] be the angle at the eye,
    [tex]tan(\theta_1)= 3/b[/tex]

    The second right triangle you should see, formed by the horizontal line from the person's eye to the wall and the line from the person's eye to the top of the picture, has horizontal base b feet and height 7 feet. If you let [itex]\theta_2[/itex] be the angle at the person's eye,
    [tex]tan(\theta_2)= 7/b[/tex]

    The angle you want is the difference between those two angles, [itex]\theta_2- \theta_1[/itex].
  5. Jan 15, 2009 #4
    Hey loop quantum gravity, you said you have a picture of what to do, uhm where is it?:redface: sorry if its a stupid question, I'm new to this!:shy:
    Thank you guys already
  6. Jan 15, 2009 #5
    Anna, my scanner isn't working.
    Anyhow you should draw 16 feet wall, where the 7 foot picture bottom is 9 feet above the ground, then the viewer is distant b feet from the wall and his height is 6 feet.
    Draw a straight line from the viewer's eyes to the pictures' bottom, and then another straight line to the top of the picture from the viewer's eyes, these two straight line are hypotenuses of two right triagnles whose sides are: (bottom) 3 and b; (up) 10 and b use twice pythagoras theorem, and then you get the sides of the triangle with the angle theta.

    You should pay attention that I haven't seen your picture, but I guess the angle theta is that between the field of vision of the viewer.
  7. Jan 15, 2009 #6
    Hi, the prof tried to give us a hint and drew the picture for us! The triangle in question is apparently NOT a right angled triangle:confused: and i am now thoroughly confused:cry:

    Attached Files:

  8. Jan 15, 2009 #7


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    Hi AnnaSuxCalc! :smile:
    Yes, you are!​

    If the top and bottom of the picture are T and B, and you are U, then the prof is correct … the angle is BUT …

    so that's the triangle the prof is talking about …

    but you don't know two of the sides of that triangle, so you have to follow HallsofIvy's :smile: plan and draw the triangles BUF and TUF (where F is the floor), and find the angles BUF and TUF. :wink:
  9. Jan 15, 2009 #8
    Hey, first of all THANKS, but also since you probably mean 9+7=16 (not 13) would the two then be:
    tan(\theta_1)= 3/b
    tan(\theta_2)= 10/b ???
  10. Jan 15, 2009 #9
    Hey, I think I know what has been really confusing me, I just understood that I am not supposed to end up with a numerical answer for theta, I just need to come up with the equation right?
  11. Jan 15, 2009 #10
    OMG, it seems soo simple but Ive just been staring at this for a while and I just cannot figure it out, I drew the triangles and I understand that the θ I need is the difference between θ2 and θ1
    I also know that tanθ=opp./adj. and we are using this because we don't know the hyp. and are supposed to write the equation in terms of b (i.e. the adj.)
    but wth do I do now?
    Should I just write tanθ=(10/b)-(3/b) this seems way to simple...do i not need to use the cos Law and get something similar to the equation quantum posted?:cry::confused:
    this is pretty embarrassing but I just can't seem to get it...
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