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Math Help - line equations, trig

  1. #1
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    line equations, trig

    The lines l1 and l2, with equations y=2x and 3y=x−1 respectively, are drawn on the same set of axes. Given that the scales are the same on both axes and that the angles that l1 and l2 make with the positive x-axis are A and B respectively,

    (a) write down the value of tan A and the value of tan B;

    (b) without using your calculator, work out the acute angle between l1 and l2.

    Really dont understand this question, how do I relate tanA and tanB with the equations?
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Really dont understand this question, how do I relate tanA and tanB with the equations?
    \tan{A} = 2

    \tan{B} = \frac{1}{3}

    \tan(A-B) = \frac{\tan{A}-\tan{B}}{1 + \tan{A}\tan{B}}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \tan{A} = 2

    \tan{B} = \frac{1}{3}

    \tan(A-B) = \frac{\tan{A}-\tan{B}}{1 + \tan{A}\tan{B}}
    Thanks, I know what the answers are but don't know why ? Could you please explain,

    Cause I thought of putting y=0 and x=0 and drawing the lines and still can't understand the answers. I can understand that to fine the acute angle, you would have to minus A from B.
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  4. #4
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    Quote Originally Posted by Tweety View Post
    Thanks, I know what the answers are but don't know why ? Could you please explain,

    Cause I thought of putting y=0 and x=0 and drawing the lines and still can't understand the answers. I can understand that to fine the acute angle, you would have to minus A from B.
    Hi Tweety,

    The tangent ratio is the same as the slope. So that's why

    \tan A=2 and \tan B=\frac{1}{3}

    As far as the acute angle (let's call it \theta), formed by the two lines, you can use the exterior angle theorem to deduce that:

    \angle A - \angle B = \angle \theta

    Follow the previous post to determine that \tan \theta = 1. Thus \angle \theta = 45^{\circ}
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  5. #5
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    Hi Tweety,

    the slope of a line can be represented by placing a right-angled
    triangle against it.

    If the line has a positive slope, place the triangle under it,
    so that the tangent of the lower left corner angle is opp/adj.
    This is the exact same way to calculate the line slope,
    using the differences between co-ordinates.

    If the line has a negative slope, use the negative of the triangle's
    tangent. So, a negative tangent means the slope of a negative-going line.

    tan(angle) = line slope.
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