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Math Help - need help on Slopes

  1. #1
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    need help on Slopes

    Find the tangents of the angles of the triangle ABC, and find the angles to the nearest degree.

    A(1,3),B(-2,-4),C(3,-2)


    i dunno how to find their angles and tans

    Thanks very much
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  2. #2
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    Do you have the idea of ‘vectors’ to work with?
    We can use the arccosine of dot products to find the angles.
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the tangents of the angles of the triangle ABC, and find the angles to the nearest degree.

    A(1,3),B(-2,-4),C(3,-2)
    i dunno how to find their angles and tans
    Thanks very much
    Hello,

    1. Calculate the length of the sides(?) of the triangle:
    |AB|=\sqrt{(1-(-2))^2+(3-(-4))^2}=\sqrt{58}

    |AC|=\sqrt{(1-3)^2+(3-(-2))^2}=\sqrt{29}

    |BC|=\sqrt{((-2)-3)^2+((-4)-(-2))^2}=\sqrt{29}

    2. As you easely can see is:

    (AC)^2+(BC)^2=(AB)^2. That means you have an isosceles right triangle. The angles at A and B are 45° and tan(45°) = 1.

    The angle at C is 90° and tan(90°) doesn't exist.

    EB
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!

    EB had the best solution . . . elegant!


    But there is a formula for this problem.

    . . Given two lines with slopes m_1 and m_2
    . . the angle \theta between them is given by: . \tan\theta \:=\:\frac{m_2 - m_1}{1 + m_1\!\cdot\!m_2}


    Find the tangents of the angles of the triangle ABC,
    and find the angles to the nearest degree.

    . . A(1,3),\;B(-2,-4),\;C(3,-2)

    We have: . \begin{array}{ccccc}m_{AB}\:=\\ \\ m_{BC}\:=\\ \\ m_{AC}\:=\end{array} \\<br />
\begin{array}{ccccc}\frac{-4-3}{-2-1} \\ \\ \frac{-2+4}{3+2} \\ \\ \frac{-2-3}{3-1} \end{array}<br />
\begin{array}{ccccc}=\;\frac{7}{3} \\ \\ =\;\frac{2}{5} \\ \\ =\,\text{-}\frac{5}{2}\end{array}


    Then: . \begin{array}{ccccc}\tan A\:=\:\frac{m_{AC} - m_{AB}}{1 + m_{AC}\cdot m_{AB}}\:= \\  \\ \tan B\:=\:\frac{m_{AB} - m_{BC}}{1 + m_{AB}\cdot m_{BC}}\:= \\ \\ \tan C \:=\:\frac{m_{AC} - m_{BC}}{1 + m_{AC}\cdot m_{BC}}\:= \end{array}<br />
\begin{array}{ccccc}\frac{-\frac{5}{2} - \frac{7}{3}}{1 + \left(-\frac{5}{2}\right)\left(\frac{7}{3}\right)} \\ \frac{\frac{7}{3} - \frac{2}{5}}{1 + \left(\frac{7}{3}\right)\left(\frac{2}{5}\right)} \\ \frac{-\frac{5}{2} - \frac{2}{5}}{1 + \left(-\frac{5}{2}\right)\left(\frac{2}{5}\right)} \end{array} \begin{array}{ccccc}=\;1 \\ \, \\=\;1\\  \\ = \;\infty\end{array}<br />
\quad\Rightarrow\quad \begin{Bmatrix}A \:= \;45^o \\ \\ B \:= \:45^o \\ \\ C \:=\:90^o\end{Bmatrix}<br />

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