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Thread: Need help with this Math Question!

  1. #1 Need help with this Math Question! 
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    I have been attempting this question for several days now! I haven't a single clue as to how to get the final answer. It's beginning to get frustrating, haha, and I really want to understand how this question is done. The question is as follows:

    Question.jpgQuestion.jpg


    The question is to find the distance between P and Q.


    Previous attempts: I have tried to solve for the arc of a smaller circle within the larger circle that is defined by a diameter of 9. This gave me an arc length of (9pi)/2. From there, I had no clue where to go with it. I also tried to calculate the radius of the circle using only the chord, but to no avail. (P.S. - I have no clue what the answer is). The only real clue I have is that this question appears in a polynomial chapter, meaning, it could have to do with solving some polynomial equation after figuring the equation out.

    Any help would be appreciated! If you could take me through this step by step, I would really appreciate it!

    Thank you in advance!

    *The picture seems to be too small. The length of the chord is 18, and the length ride beside the chord (the other line that seems to be making a triangle with the top line) has a length of 26.
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  2. #2  
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    It would be helpful if you started from the beginning. What is the original question?
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  3. #3  
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    Oops! Sorry! The question is to find the distance between P and Q.
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  4. #4  
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    What is the significance of the green arrow?
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  5. #5  
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    Quote Originally Posted by Jilan View Post
    What is the significance of the green arrow?
    It's just showing that the line going down the middle of the circle is the radius
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  6. #6  
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    OK, I can see how you can get it in terms of the radius. Let's call it r. Can you see that the cosine of the angle QPB is 18/2r?
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    Quote Originally Posted by Jilan View Post
    OK, you can only get PQ in terms of the radius. Let's call it r. Can you see that the cosine of the angle QPB is 18/2r?
    Ahh! Yes I see that now! What happens next?
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  8. #8  
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    Quote Originally Posted by Jilan View Post
    OK, I can see how you can get it in terms of the radius. Let's call it r. Can you see that the cosine of the angle QPB is 18/2r?
    Yes, I see it now
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  9. #9  
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    Try drawing a line vertically up from point B and call it T. Then PQ= PT + r.
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  10. #10  
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    Quote Originally Posted by Jilan View Post
    Try drawing a line vertically up from point B and call it T. Then PQ= PT + r.
    Wow, how smart lol! I've done that, what next?
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  11. #11  
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    You should now also be able to to see how to get r, but I don't want to spoil it for you
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  12. #12  
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    Quote Originally Posted by Jilan View Post
    You should now also be able to to see how to get r, but I don't want to spoil it for you
    I still can't get it I tried
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  13. #13  
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    OK, sin QPB = r/26 and cos QPB = 18/2r. Does that help?
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  14. #14  
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    Quote Originally Posted by Jilan View Post
    OK, sin QPB = r/26 and cos QPB = 18/2r. Does that help?
    A bit. Could you give me a few more hints please and is QPB angle P in triangle QPB?
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    Quote Originally Posted by BitRunner View Post
    A bit. Could you give me a few more hints please and is QPB angle P in triangle QPB?
    Yes the angle QPB is the angle between the lines QP and PB, the pointy one at P. You then need to use the identity sin ^2 +cos ^2 =1 to solve for r.
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  16. #16  
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    Quote Originally Posted by Jilan View Post
    Yes the angle QPB is the angle between the lines QP and PB, the pointy one at P. You then need to use the identity sin ^2 +cos ^2 =1 to solve for r.
    Okay, so how would I set up the values in the identity? No matter what I did, I couldn't see what to do.
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  17. #17  
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    See post #13.
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  18. #18  
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    Quote Originally Posted by Jilan View Post
    See post #13.
    (r/26)^2 + (18/2r)^2 = 1?
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    Looking good.
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  20. #20  
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    Quote Originally Posted by Jilan View Post
    Looking good.
    And that should give me r?

    One question. Triangle QPB is not a right triangle. Why is it that sin of QPB and cos of QPB give those values?
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    Remember the point T in post #9 ? The triangle BTP is a right angled triangle. I am assuming that BT = r as it looks like it ought to be. But that would be assuming that the green arrow has more significance than you said, so I am still unsure whether this will work or not
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    Quote Originally Posted by Jilan View Post
    Remember the point T in post #9 ? The triangle BTP is a right angled triangle. I am assuming that BT = r as it looks like it ought to be. But that would be assuming that the green arrow has more significance than you said, so I am still unsure whether this will work or not
    The green arrow has no more significance other than what I pointed out. The original diagram did not have the green arrow in the diagram. I drew the arrow because the diagram was too small.
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  23. #23  
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    Fair enough. Lets assume BOQ is a right angle and keep going. Are you OK with solving for r now?
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    Maybe this may help. This ain't mine, but I am learning on it. Distances and Sizes

    One way to measure distances is if you know the size of an object. If you can then measure its apparent angular size, you can work out the distance:




    Given the geometry above, and if D≫r, (almost always the case in astronomy), then

    r=θD
    as long as θ is measured in radians (a radian is 180π degrees).

    Angles in astrophysics are often so small that even a degree is too large a unit to be convenient. We typically use arcminutes (one arcminute is 1/60 of a degree) and arcseconds (one arcsecond = 1/60 of an arcminute = 1/3600 of a degree).


    Fluxes and luminosities

    Luminosity L is the total amount of power put out by some object, and is measured in Watts. Flux f is the power we receive at our telescope, per unit collecting area, and is measured in Watts per square metre. They are related by the equation:

    f=L4πD2
    where D is the distance to the emitting object.


    Spectra and the Doppler Effect

    A spectrum is a graph of flux per unit wavelength plotted against wavelength. It will often show emission or absorption lines due to particular elements.

    If an object is moving towards or away from you, these spectral lines will be moved in wavelength away from their normal wavelength λo. If you observe a line at wavelength λ, you can define a redshift z as:

    z=λ−λ0λ0
    If this shift is due to the doppler effect, and the velocity v<<c (velocity much less than the speed of light - nearly always true), then:

    z=vc
    To measure a redshift, you will need to know what lines to expect, and what their wavelengths are in the laboratory. The following graph shows you some of the typical lines you would see in a star or galaxy. Note that not all stars will show all these lines, and there are a variety of other lines that in certain stars can be strong. The C-H line is due to vibrations in the chemical bond linking carbon to hydrogen in molecules.



    Hubble Law

    Assuming that the brightest star in every galaxy had about the same luminosity (not a good assumption), Edwin Hubble calculated their relative distances. He found that the distances correlated with redshift. Everything was moving away from us and the speed correlated with how far things were from us.

    The standard explanation is that space itself is expanding. Objects are not moving - they are just being carried apart by the expansion of space.

    This means that unless more matter is created, the density of the universe must continuously go down (same amount of matter spread over more space). The alternative is that more matter is appearing out of nowhere - this is called the “steady state theory”.

    The “Steady State Theory” predicts that the universe should always look the same. We actually observe, however, that the universe was different in the past (we can see the past by looking a distant objects). Quasars were more common and the microwave background emission comes from a time when space was opaque.
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  25. #25  
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    Quote Originally Posted by Jilan View Post
    Fair enough. Lets assume BOQ is a right angle and keep going. Are you OK with solving for r now?
    I'm still not too sure. If it is possible, could you show me how you would take this step by step? I understand that BOQ is the right triangle; are the rations still the same (of sin and cos)? In any case, only if possible, could you show me the steps you would take?
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  26. #26  
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    Quote Originally Posted by mayflow View Post
    Maybe this may help. This ain't mine, but I am learning on it. Distances and Sizes

    One way to measure distances is if you know the size of an object. If you can then measure its apparent angular size, you can work out the distance:




    Given the geometry above, and if D≫r, (almost always the case in astronomy), then

    r=θD
    as long as θ is measured in radians (a radian is 180π degrees).

    Angles in astrophysics are often so small that even a degree is too large a unit to be convenient. We typically use arcminutes (one arcminute is 1/60 of a degree) and arcseconds (one arcsecond = 1/60 of an arcminute = 1/3600 of a degree).


    Fluxes and luminosities

    Luminosity L is the total amount of power put out by some object, and is measured in Watts. Flux f is the power we receive at our telescope, per unit collecting area, and is measured in Watts per square metre. They are related by the equation:

    f=L4πD2
    where D is the distance to the emitting object.


    Spectra and the Doppler Effect

    A spectrum is a graph of flux per unit wavelength plotted against wavelength. It will often show emission or absorption lines due to particular elements.

    If an object is moving towards or away from you, these spectral lines will be moved in wavelength away from their normal wavelength λo. If you observe a line at wavelength λ, you can define a redshift z as:

    z=λ−λ0λ0
    If this shift is due to the doppler effect, and the velocity v<<c (velocity much less than the speed of light - nearly always true), then:

    z=vc
    To measure a redshift, you will need to know what lines to expect, and what their wavelengths are in the laboratory. The following graph shows you some of the typical lines you would see in a star or galaxy. Note that not all stars will show all these lines, and there are a variety of other lines that in certain stars can be strong. The C-H line is due to vibrations in the chemical bond linking carbon to hydrogen in molecules.



    Hubble Law

    Assuming that the brightest star in every galaxy had about the same luminosity (not a good assumption), Edwin Hubble calculated their relative distances. He found that the distances correlated with redshift. Everything was moving away from us and the speed correlated with how far things were from us.

    The standard explanation is that space itself is expanding. Objects are not moving - they are just being carried apart by the expansion of space.

    This means that unless more matter is created, the density of the universe must continuously go down (same amount of matter spread over more space). The alternative is that more matter is appearing out of nowhere - this is called the “steady state theory”.

    The “Steady State Theory” predicts that the universe should always look the same. We actually observe, however, that the universe was different in the past (we can see the past by looking a distant objects). Quasars were more common and the microwave background emission comes from a time when space was opaque.
    I doubt this has any relevance to the question. It's an advanced functions (12) question; not this advanced though lol
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  27. #27  
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    image.jpg
    Sorry it's so messy.
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  28. #28  
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    Quote Originally Posted by Jilan View Post
    image.jpg
    Sorry it's so messy.
    According to my teacher, that is the correct answer :O! (Even my teacher couldn't get it!)

    One question though, I followed up until near the end of the calculation using the quadratic formula. How did you get 94 and 9.7? And I couldn't read the stuff on the top right of the paper (the sin theta = and the the calculation to get PQ).
    Could you type that out for me?

    On another note, thank you so much! You've been a big help! Really appreciate it. And lol, don't worry, my work is even messier :P
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  29. #29  
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    From the quadratic formula you get
    r^2 = 338 + 244 or
    r^2 = 338 - 244
    I picked the second one as the first one gives an r that looks too big.
    So r^2 = 94 and r = sqrt 94 = 9.7.

    From triangle BTP
    TP^2 + BT^2 = 26^2
    TP^2 = 26^2 - BT^2
    and assuming that BT = r (the bit we were not sure about)
    TP^2 = 676 -94 = 582
    TP = 24.1

    PQ = PT + QT = 24.1 + 9.7 = 33.8
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  30. #30  
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    I can always do these when the information required is there.
    Which it is not.
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  31. #31  
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    Quote Originally Posted by pikpobedy View Post
    I can always do these when the information required is there.
    Which it is not.
    Life is often a bit like that I find. The data is often incomplete, so we work with what we're given and state the assumptions. But we never give up!
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  32. #32  
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    Quote Originally Posted by Jilan View Post

    From the quadratic formula you get
    r^2 = 338 + 244 or
    r^2 = 338 - 244
    I picked the second one as the first one gives an r that looks too big.
    So r^2 = 94 and r = sqrt 94 = 9.7.

    From triangle BTP
    TP^2 + BT^2 = 26^2
    TP^2 = 26^2 - BT^2
    and assuming that BT = r (the bit we were not sure about)
    TP^2 = 676 -94 = 582
    TP = 24.1

    PQ = PT + QT = 24.1 + 9.7 = 33.8
    THANK YOU SO MUCH! It all makes sense now! Really really appreciate your help! Thanks a bunch!
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  33. #33  
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    Quote Originally Posted by Jilan View Post
    Life is often a bit like that I find. The data is often incomplete, so we work with what we're given and state the assumptions. But we never give up!
    Math immitating life. Usually engineering and physics do that while math prides itself in being exact.

    When drawings are eyeballed and assumptions are made and explicitly stated as such, I agree. It "looks" to scale, equal, right angle, 45 degrees, parallel and so forth.

    The result can be differing correct answers. The question becomes is there a best answer in the choices given or is the student marked on the total explocit calculation work?
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  34. #34  
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    Maybe marks should be given for the number of pieces of paper that get used.
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