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Math Help - Square and triangle.

  1. #1
    Newbie Sorombo's Avatar
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    Square and triangle.

    There is a ABCD square, inside him there is a P point. The angle BAP=angle ABP= 15 . Find the angle PCD.

    Note: Do not use trigonometry.
    Attached Thumbnails Attached Thumbnails Square and triangle.-problema-cuadrado-triangulo.png  
    Last edited by Sorombo; September 19th 2011 at 01:45 PM.
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  2. #2
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    Re: Square and triangle.

    Here's a kicker,

    BPC = APD = 90-15 = 75.

    APB = 180 - (15+15) then you should be able to find CPD.
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  3. #3
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    Re: Square and triangle.

    Can you please explain how you reason out 'BPC = APD = 90-15 = 75. '.
    Thanks
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  4. #4
    Member anonimnystefy's Avatar
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    Re: Square and triangle.

    hi pickslides

    shouldn't PBC=75 and not BPC.
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  5. #5
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    Re: Square and triangle.

    Thanks can you explain how the above facts allow x to be found.
    Thanks
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  6. #6
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    Re: Square and triangle.

    P is a point on the perpendicular bisector ofAB and DC
    Altitude of APB =tan15*.5 whereAB=1
    Altitude of DPC is 1-tan15*.5 =.866025
    sin x = .866025 x=60
    Last edited by bjhopper; September 19th 2011 at 01:01 PM. Reason: correction
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  7. #7
    Newbie Sorombo's Avatar
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    Re: Square and triangle.

    Quote Originally Posted by pickslides View Post
    Here's a kicker,

    BPC = APD = 90-15 = 75.

    APB = 180 - (15+15) then you should be able to find CPD.

    I don't understand why APD=90-15

    could u explain it? i didn't get the argument
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  8. #8
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    Re: Square and triangle.

    Hi guys, I meant PAD = 75
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  9. #9
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    Re: Square and triangle.

    If you cannot use trig assume x=60 and work back showing that angles PAE and PEA equal 15.Any other angle for x will produce different ones
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  10. #10
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    Re: Square and triangle.

    Using the fact that triangles DPC and ABP are isoseles label all missing angles and set up a system of equations to solve.

    We know that ABC = BAD = 75 and APB = 150.

    Labelling APD = BPC = a and BCP = ADP = y and CDP = DCP = x and DPC = b

    You should have 4 equations with four unknowns.

    x+y = 90
    2x+b = 180
    2a+b+150=360
    y+a+75=180

    Now solve the system.
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  11. #11
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    Re: Square and triangle.

    There is another way. Draw two circles with a radius side of square fromC and D .ADP is then isosceles
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  12. #12
    Newbie Sorombo's Avatar
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    Re: Square and triangle.

    Quote Originally Posted by pickslides View Post
    Using the fact that triangles DPC and ABP are isoseles label all missing angles and set up a system of equations to solve.

    We know that ABC = BAD = 75 and APB = 150.

    Labelling APD = BPC = a and BCP = ADP = y and CDP = DCP = x and DPC = b

    You should have 4 equations with four unknowns.

    x+y = 90
    2x+b = 180
    2a+b+150=360
    y+a+75=180

    Now solve the system.
    I think the system can't be solved. Because the 4th equation come from 1/2*(3rd-2nd)+1st
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  13. #13
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    Re: Square and triangle.

    Hello Sorombo,

    Did you try drawing the two circles.If you did you should see in addition to the two isosceles triangles ADP and CDE that the intersections of the circles meet along the perpendicular bisector of CD and AB. It follows that DCP is equilateral
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  14. #14
    Newbie Sorombo's Avatar
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    Re: Square and triangle.

    Quote Originally Posted by bjhopper View Post
    Hello Sorombo,

    Did you try drawing the two circles.If you did you should see in addition to the two isosceles triangles ADP and CDE that the intersections of the circles meet along the perpendicular bisector of CD and AB. It follows that DCP is equilateral
    I was thinking about the 2 circles, but i don't know how you know that triangle ADP is isosceles. In fact, I think I must show that point P belongs to the circle.
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  15. #15
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    Re: Square and triangle.

    Sorobam,
    Choose a length of the side of square as a radius and draw the two circles.Connect the two circle intersections. This line is the perpendicular bisector of AB and DC. ADP is isosceles because each of its long sides is a radius but also DC, DP, and CP are radii so DPC is equilateral.
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