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Thread: Matlab Trigonometry Question

  1. #1
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    Matlab Trigonometry Question

    Matlab Trigonometry Question-untitled.png

    https://i.imgur.com/isLpH5w.png

    x = [20:6:68];
    t1=atan((8-x.*tan(8*pi/180))/x);
    t=atan((22-x.*tan(8*pi/180))/x)-t1;
    Table = [t1' t'];
    disp(Table)
    Not entirely sure if its correct. Need help...
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  2. #2
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    Re: Matlab Trigonometry Question

    Let the positive x-axis be to the left.

    The top of the screen is at $p_{t}=(0,30)$
    The bottom of the screen at $p_{b}=(0,8)$

    A point on the sloped line at distance $x$ from the screen is at $p=(x, x\tan(8^\circ))$

    A vector from the point $p$ to the top of the screen is $v_t=p_t-p=(-x,30-x\tan(8^\circ))$
    A vector from the point $p$ to the bottom of the screen is $v_b=p_b - p = (-x, 8-x\tan(8^\circ))$

    The angle between these two vectors is $\theta = \arccos\left(\dfrac{v_t \cdot v_b}{|v_t||v_b|}\right)$

    This produces different results than your formula.

    Rewrite your program to perform the above.
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