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Math Help - Tangents and working angles...

  1. #1
    Senior Member Mukilab's Avatar
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    Tangents and working angles...

    I'll try to draw a picture in your minds, I seem to not be able to solve it at all and I need it fast :L

    There is a circle. It has a polygon inside of it, a bit shaped like a boomerang. This polygon has 4 edges. 3 of the verts are touching the circle. point A is at the bottom and slightly to the right. B is at the top and less to the right than A. C is to the left. O is the centre. the angle of point C of the polygon (inner) is 75 degrees.

    There are also two tangents, diagonal lines extending form point P which is to the right. These touch points A and B (bottom and top).

    Work out AOB (inner of O) and a reason (please show me a method and Work In Progress)

    Then work out the angle of APB, the inner angle of the two tangents meeting at point P.


    Sorry for not being able to scan.
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  2. #2
    Super Member Quacky's Avatar
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    Even though I think you have described it well, I think I'll struggle to correctly visualise the diagram. Try reading through this page, however, as the theorems are all there:

    Circle Theorems
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  3. #3
    Senior Member Mukilab's Avatar
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    Wow it retains exactly what I need *hurries to write more formulas*

    I'll have pages by the end, thank you
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  4. #4
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    Hello, Mukilab!

    I don't know you mean by "the angle of point C (inner)".
    But I'll take a guess.


    I'll try to draw a picture in your minds.

    There is a circle. It has a polygon inside of it, a bit shaped like a boomerang.
    This polygon has 4 edges. 3 of the verts are touching the circle.
    Point A is at the bottom and slightly to the right.
    B is at the top and less to the right than A.
    C is to the left. O is the centre.
    The angle of point C of the polygon (inner) is 75.

    There are also two tangents, diagonal lines extending from point P which is to the right.
    These touch points A and B (bottom and top).

    Work out angle AOB.

    Then work out angle APB.
    Code:
                  * * *
              *           *  B
            *               o
           *              *   *
                        *       * 
        C o           *       *   *
          *       O *         *     o P
          *           *       *   *
                        *       *
           *              *  *
            *               o
              *           *  A
                  * * *

    Draw chords CA and CB.

    We are told: . \angle BCA \,=\,75^o

    . . Hence: . \text{arc}(AB) \:=\:150^o

    Therefore: . \boxed{\angle AOB \:=\:150^o}


    In quadrilateral BPAO:\;\;\angle APB + \angle PAO + \angle AOB + \angle OBP \:=\:360^o

    . . \angle APB + 90^o + 150^o + 90^o \:=\:360^o \quad\Rightarrow\quad \boxed{\angle APB \:=\:30^o}

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  5. #5
    MHF Contributor
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    The boomerang is made of 2 isosceles triangles,
    because the centre of the circle is the "inside" vertex of the boomerang.

    The triangles are isosceles due to 2 sides being the radius length.
    If a triangle is isosceles, then the angles opposite the sides of equal length
    are equal (an isosceles triangle is made of 2 identical back-to-back
    right-angle triangles).

    All 3 angles in a triangle sum to 180 degrees.
    The angles in a semicircle (or the 2 angles on one side of a line) sum to 180 degrees.
    The angles in a circle sum to 360 degrees.

    All of these geometric truths solve these types of question.

    X+Y=75^o

    W=180^o-2X

    V=180^o-2Y

    For angle AOB, 180-W+180-V=360-(W+V)=360-(360-[2X+2Y])

    =2(X+Y)=2(75^o)=150^o

    A well-known result is...
    In a circle, the angle at the centre is twice the angle at the circumference
    for 2 angles standing on a common arc.
    The common arc here is the arc from A to B on the right, just to the left of P.

    Triangles oAP and oBP are identical as they have 2 sides the same length
    (green dotted hypotenuse and R) and both have a 90 degree angle due to the tangents.
    Therefore they are identical right-angled triangles.

    The sum of the angles of 2 triangles is 180+180=360 degrees.
    Hence, angle APB=360-(90+90+180-W+180-V) or 360-(180+150)=30 degrees.
    Attached Thumbnails Attached Thumbnails Tangents and working angles...-boomerang.jpg  
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  6. #6
    Senior Member Mukilab's Avatar
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    Edit: I do not understand how it is made of two isosceles triangles.
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  7. #7
    Senior Member Mukilab's Avatar
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    Sorry, yesterday I did not have time to view the circle theorem link, only skim and today now I look over it I have solved my first question. Quite easily actually, I apologise but still thank both of you for taking the consideration to explain such a simplistic method to such a simplistic person.

    I know need to know the angle of the place where the tangents meet, P (the inner section). This is neither covered int the circle theorem link nor in neither of those posts. Any further help? Sorry for the inconvenience.
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  8. #8
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    Hi Mukilab,

    If you move A or B to make an non-symmetrical boomerang,
    the two boomerang parts (if you break it along the line oC)
    will still be isosceles and the maths still follows.

    That is if A, B and C are on the circle circumference and if o is the centre.
    You can rotate the points A, B and C around the circumference and the 2 triangles of the boomerang will still be isosceles, though different sizes.

    In the sketch, the angles X and Y can vary, but both isosceles triangles will still individually have 2 angles the same size.

    The only way the triangles would not be isosceles is if any of A, B, C were off the circumference, or if o was not the centre, or if the shape was not a circle.

    This problem covers a number of geometric themes,
    we can resolve what you need to understand about it yet.

    I will send another sketch later.
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  9. #9
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Hi Mukilab,

    If you move A or B to make an non-symmetrical boomerang,
    the two boomerang parts (if you break it along the line oC)
    will still be isosceles and the maths still follows.

    That is if A, B and C are on the circle circumference and if o is the centre.
    You can rotate the points A, B and C around the circumference and the 2 triangles of the boomerang will still be isosceles, though different sizes.

    In the sketch, the angles X and Y can vary, but both isosceles triangles will still individually have 2 angles the same size.

    The only way the triangles would not be isosceles is if any of A, B, C were off the circumference, or if o was not the centre, or if the shape was not a circle.

    This problem covers a number of geometric themes,
    we can resolve what you need to understand about it yet.

    I will send another sketch later.

    Yes, sorry I have just realised this. Please read my update though.
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  10. #10
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    Hi Mukilab,

    Angle APB is the angle at P (inner angle).
    APB means start at A, go to P and end up at B.
    Angle BPA is the same angle.

    The attachments show a bit on isosceles triangles, tangents and the angle at P.
    Attached Thumbnails Attached Thumbnails Tangents and working angles...-tangent.jpg   Tangents and working angles...-isosceles1.jpg   Tangents and working angles...-isosceles2.jpg  
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