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Math Help - Angles... :(

  1. #1
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    Angles... :(

    In the figure, ABCD is a square and ABE is an equilateral triangle. Find <ADE. I just for the life of me find out what E and D are... I got A, but... Grrrr...
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  2. #2
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    And then another question... I don't get where my teacher gets this stuff...

    (x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

    Any and all help is appreciated!
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  3. #3
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    Quote Originally Posted by Rocher View Post
    In the figure, ABCD is a square and ABE is an equilateral triangle. Find <ADE. I just for the life of me find out what E and D are... I got A, but... Grrrr...
    AB = AE because triangle AEB is equilateral.

    AD = AB because ABCD is a square.

    AD = AE because AD = AB = AE.

    Thus triangle EAD is isosceles.

    Thus angle ADE = angle AED.

    Angle DAE is the complement of angle EAB because the interior angles of a square are right angles.

    Angle EAB is 60 degrees because triangle AEB is equilateral.

    Thus angle DAE is 90 - 60 = 30 degrees.

    Call angle AED x. Then 30 + x + x = 180 because the sum of the interior angles of a triangle is 180 degrees.

    Solving this equation I get that x = 75, so angle AED = angle ADE= 75 degrees.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    And then another question... I don't get where my teacher gets this stuff...

    (x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4) is an identity in x, find the values of A and B.

    Any and all help is appreciated!
    This really should have gone in its own thread. See rule #15 here.

     (x-4)(x-2)=(x-3)(x-5)-A(x-3)-B(x-4)

    Expanding
    x^2 - 6x + 8 = x^2 - 8x + 15 - Ax + 3A - Bx + 4B

    Collecting like terms:
    (-6 + 8 + A + B)x + (8 - 15 - 3A - 4B) = 0

    (A + B + 2)x + (-3A - 4B - 7) = 0

    In order for this expression to be 0 for all x we must have the coefficients of all powers of x equal to 0. So
    A + B + 2 = 0
    and
    -3A - 4B - 7 = 0

    Can you take things from here?

    -Dan
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