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MATH

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FIND THE SOLUTION FOR THE EQUATION. ASSUME THAT ALL ANGLES IN WHICH AN UNKNOWN APPEARS ARE ACUTE ANGLES.

COT ALPHA=TAN( ALPHA +40DEGREES)


HOW DO I SOLVE THIS?

  • MATH -

    I will use A for alpha

    cot A = tan(A + 40)
    1/tanA = (tanA + tan40)/(1 - tan^2A)
    let tanA = x

    1/x = (x + .8391)/(1-x(.8391))
    x^2 + .8391x = 1 - .8391x
    x^2 + 1.6782x - 1 = 0
    by the formula
    x = (-1.6782 ± √6.816355)/2 = .4663 or -2.1445

    so tanA = .4663, then A = 25° or 205°
    if tanA = -2.1445 , A = 115° or 295°

    I will check the 205°
    LS = cot 205° = 2.1445
    RS = tan(205+40) = tan 245 = 2.1445
    YEAHHH!

    I will let you check the others.

  • MATH -

    Just came back from a walk, and I kept thinking about this question and why the answer came out to rather exact values of the angle.
    Then I realized that I could have used the complementary angle property, that is,
    cot A = tan(90°-A)

    so in
    cotA = tan(A+40)
    tan(90-A) = tan(A+40)
    so obviously
    90-A = A + 40
    50 = 2A

    A = 25 !!!!!!!!!!!!

    and of course since tanA = tan(180-A)
    A is also equal to 180-25 = 115°

    and since the period of tanA is 90° , adding or subtracting multiples of 90 would yield more answers
    so 25+90 = 115
    115+90 = 205
    205+90 = 295

    which is the same answer as the complicated solution I had before.

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