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Math Help - maths test questions

  1. #1
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    maths test questions

    I'm really stuck practicing for my test and i cant get these questions.
    Please show working out.

    1. Show that the triangle whose vertices are (0,1), (2,-1) and (-sqrt(3)+1,-sqrt(3)) is equilateral.


    2. ImageShack - Hosting :: mathxw9.jpg
    I'm sorry it is upside down


    Thanks
    Last edited by poniekid; April 26th 2008 at 11:19 PM. Reason: change questions
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  2. #2
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    Hello,

    For 1), calculate the norm of the vectors joining two points... do it 3 times

    If A(x_A;y_A) and B(x_B;y_B), then :

    ||\vec{AB}||=\underbrace{AB}_{\text{distance AB}}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}

    This formula can be retrieved by using the Pythagorean theorem while dealing with a graph.

    So let's take an example :

    A(0;1) and C(-\sqrt{3}+1,-\sqrt{3})

    \begin{array}{ccc} AC & = & \sqrt{(0-(-\sqrt{3}+1))^2+(1-(-\sqrt{3}))^2} \\ & = & \sqrt{(\sqrt{3}-1)^2+(1+\sqrt{3})^2} \\ & = & \sqrt{3-2\sqrt{3}+1+1+2\sqrt{3}+3} \\ & = & \sqrt{8} \\ & = & 2 \sqrt{2} \end{array}

    Do the same for the others
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  3. #3
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    For number 2:

    Use this formula:

    Let acute angle between two lines =  x
     tan x = |\frac{m_1-m_2}{1+m_1m_2}|

    where m_1 the gradient of one of the lines, and  m_2 is the gradient of the other line. (1 and  \frac{1}{\sqrt3} in your case)
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  4. #4
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    Quote Originally Posted by poniekid View Post
    ...

    2. ImageShack - Hosting :: mathxw9.jpg
    I'm sorry it is upside down
    to #2:

    Let a denote the angle which is included by a straight line and the x-axis. Then the slope m (or gradient) of the line is:

    \tan(a) = m

    If \tan(a) = 1~\implies~a=45^\circ

    If \tan(a) = \frac1{\sqrt{3}}~\implies~a=30^\circ

    I've attached your drawing (not upside down!) with the results given above.

    I'll leave the rest for you.
    Attached Thumbnails Attached Thumbnails maths test questions-achse_gerade_winkel.gif  
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