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Thread: Trigonometry [please i need help!]

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    Trigonometry [please i need help!]

    Trigonometry

    Q1 : Given that tan 2y = -0.7265, find all the values of y between 0 degree and 360 degree.

    Q2: attached picture. [P1010892]

    Solution of Triangles

    Q1: attached picture. [P1010893]
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    Last edited by tecm0; Jul 10th, 2006 at 01:31 AM.

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    Resolved Re: New to Trigonometry / Solution of Triangles. [HELP!]

    please help........

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    Smitten by reality Harsh Gupta's Avatar
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    Re: New to Trigonometry / Solution of Triangles. [HELP!]

    for trig, Q2, (i am using x instead of theta)

    we know that: sin2x + cos2x = 1

    putting value of sinx: 25/169 + cos2x = 1

    therefore: cos2x = 1 - 25/169
    ..............cos2x = 144/169
    ..............cosx = 12/13

    and: secx = 1/cosx
    so: secx = 13/12

    and: cotx = cosx/sinx
    so: cotx = 12/13 * -13/5 = -12/5

    the first trig question is a bit confusing, atleast to me!!
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    Re: Trigonometry [please i need help!]

    thx for it.

    man, i still need help on those 2 questions

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    Re: Trigonometry [please i need help!]

    you heard of the double angle formulas?
    Rich

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    Re: Trigonometry [please i need help!]

    2y = arctan(-0.7265) = -35.99840...

    But the tangent function is periodic, so arctan(-0.7265) also equals 144.00159..., 324.00159..., 504.00159..., 684.00159... .

    But this equals 2y, so halve all the values to get y:

    y = 72.0, 162, 252 and 342 degrees to 3 s.f. in the range 0<y<360

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    Re: Trigonometry [please i need help!]

    For your other question, use the cosine rule 3 times and simply state the largest angle you find. If the sides a,b and c are opposite angles A, B and C:

    a² = b² + c² - 2bc.cosA
    b² = a² + c² - 2ac.cosB
    c² = a² + b² - 2ab.cosC

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    Smitten by reality Harsh Gupta's Avatar
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    Re: Trigonometry [please i need help!]

    the second question, finding the angle.

    well there are 2 methods, first one is using the Laws of Sines which states, that the length of each side is proportional to the SINE of the angle opposite it. but i am unable to calculate from it. maybe i am doing something wrong.

    another method would be, use Hero's formula to find the area of the triangle, and then...

    let me give you an example
    Code:
    Hero's Formula
    ---------------------
    K = sqrt(s*[s-a]*[s-b]*[s-c]),  where  s = (a+b+c)/2
    
    so according to your question, s = 11/2 or 5.5 and K = 15.2 or 3.8
    
    also Area of the triangle is also given by:
    
    K = (1/2)*a*b*sin(C),
    
    therefore, sin(C) = 2*K/(a*b)
    
    Similarly
    
    sin(A) = 2*K/(b*c),
    sin(B) = 2*K/(c*a)
    hope it helps you.

    EDIT: here a,b,c are the sides and A,B,C are the angles opposite to a,b,c respectively.
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