1. Joined
    04 Nov '07
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    335
    26 Feb '08 07:19
    Originally posted by smaia
    You will be amazed how beautiful and simple the solution is.
    No need to use calculators or perform complex computations.
    It can be resolved mentally if you discover a few hidden things.
    I got about 19.85 degrees, but i want to find the nice simple way
    (there are omitted degrees symbols....)

    let x = AB = BC = AC
    BD = DC = x/2

    using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg

    ED/sin(60) = (x/2)/sin(50) (sine rule)
    ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))

    sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rule, 110 deg from basic trig)
    EC = sqroot(3sqr(x)/(4sqr(sin(50))) + sqr(x)/4 - 2 * sqroot(3)*sqr(x)/(4sin(50)) * cos(110)) = 1.39x

    DC/sin(DEC) = EC/sin(110)
    sin(DEC) = x/2 * sin(110)/EC
    DEC = 19.849404914... degrees

    very messy....

    correct me if im wrong please
  2. Joined
    07 Sep '05
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    35068
    26 Feb '08 14:52
    Originally posted by smaia
    If you want to follow your approach, you need to set AB = x where x is any real number.
    No you don't - the angles will be the same for any similar triangle. You can pick whatever value is most convenient. 1, for instance.
  3. R
    Standard memberRemoved
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    8528
    26 Feb '08 18:12
    Originally posted by sb3700
    I got about 19.85 degrees, but i want to find the nice simple way
    (there are omitted degrees symbols....)

    let x = AB = BC = AC
    BD = DC = x/2

    using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg

    ED/sin(60) = (x/2)/sin(50) (sine rule)
    ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))

    sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rul ...[text shortened]... * sin(110)/EC
    DEC = 19.849404914... degrees

    very messy....

    correct me if im wrong please
    can you resolve that mentally with no need for complex calculations, as the poster of this problem said the solution is simple ( manageable in the mind). If you can do all of that in your head, your the man. if i had to guess, your solution is not of the simple variety, just a thought😵
  4. Joined
    14 Feb '04
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    40742
    26 Feb '08 20:391 edit
    Angle DEC is 50 degrees, break down all the triangles within the triangle.
  5. Joined
    09 Aug '06
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    5363
    27 Feb '08 03:01
    Originally posted by mtthw
    No you don't - the angles will be the same for any similar triangle. You can pick whatever value is most convenient. 1, for instance.
    True, but assume you are not allowed to use a calculator.
  6. Joined
    09 Aug '06
    Moves
    5363
    27 Feb '08 03:07
    Originally posted by sb3700
    I got about 19.85 degrees, but i want to find the nice simple way
    (there are omitted degrees symbols....)

    let x = AB = BC = AC
    BD = DC = x/2

    using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg

    ED/sin(60) = (x/2)/sin(50) (sine rule)
    ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))

    sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rul ...[text shortened]... * sin(110)/EC
    DEC = 19.849404914... degrees

    very messy....

    correct me if im wrong please
    Nice try, but sorry, this is not the correct answer. You don't need to develop such amazingly complicated calculations to solve this problem.
    I assure you: all you need is
    1- the sum of the internal angles of a triangle is 180
    2- The properties of similar triangles
    3- the opposite angles of two intersecting lines are equal and other axioms of euclidian geometry

    Good luck.
  7. Joined
    20 Sep '07
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    987
    27 Feb '08 04:441 edit
    *edit* nevermind
  8. In Christ
    Joined
    30 Apr '07
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    172
    27 Feb '08 08:451 edit
    ...
  9. Joined
    04 Nov '07
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    335
    27 Feb '08 08:52
    hence the:
    " but i want to find the nice simple way"

    similar triangles hey......
  10. Joined
    04 Nov '07
    Moves
    335
    27 Feb '08 09:20
    has it been 2 weeks yet....
  11. Joined
    07 Sep '05
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    35068
    27 Feb '08 09:571 edit
    Originally posted by smaia
    True, but assume you are not allowed to use a calculator.
    Again, irrelevant. It's about similifying your expressions.

    I got (first attempt - will check it later):

    DEC = arctan[sin(50)sin(70)/(sin(50)cos(70) + sin(60))]

    No calculators in sight. Clearly not the 'simple' method you're looking for, though.
  12. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
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    20443
    27 Feb '08 11:28
    From the AD segment, draw a 20 degrees line that intersects AB at a point E.
    As I understand it there is no single answer.


    The quotation above has no restrictions on choosing the point E.

    You can manage it so that point E is point B, hence the requested angle is 0.

    What am I missing?
  13. Joined
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    35068
    27 Feb '08 11:40
    Originally posted by TheMaster37
    What am I missing?
    You're missing his clarification in a later post 🙂
  14. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
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    20443
    27 Feb '08 13:41
    Originally posted by mtthw
    You're missing his clarification in a later post 🙂
    Ah, shame that I should miss that post :p
  15. Backside of desert
    Joined
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    14828
    27 Feb '08 16:432 edits
    Originally posted by mtthw
    Again, irrelevant. It's about similifying your expressions.

    I got (first attempt - will check it later):

    DEC = arctan[sin(50)sin(70)/(sin(50)cos(70) + sin(60))]

    No calculators in sight. Clearly not the 'simple' method you're looking for, though.
    I came up with:

    Angle CED = arccot { [ 2 (sin 60) - (sin 110) (cos 50) ] / [ (sin 110) (sin 50) ] }

    which is equiivent with yours and smasia's first post

    appx. = 32.544

    if three of us agree then maybe his solution is flawed
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