Results 1 to 4 of 4

Thread: need help on Slopes

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    need help on Slopes

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

    $\displaystyle A(1,3),B(-2,-4),C(3,-2)$


    i dunno how to find their angles and tans

    Thanks very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1
    Do you have the idea of ‘vectors’ to work with?
    We can use the arccosine of dot products to find the angles.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the tangents of the angles of the triangle $\displaystyle ABC$, and find the angles to the nearest degree.

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

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

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

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

    2. As you easely can see is:

    $\displaystyle (AC)^2+(BC)^2=(AB)^2$. 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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, ^_^Engineer_Adam^_^!

    EB had the best solution . . . elegant!


    But there is a formula for this problem.

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


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

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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. slopes
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 2nd 2008, 09:36 AM
  2. Slopes of triangles???
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Dec 27th 2007, 07:22 PM
  3. help with slopes!
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: May 3rd 2007, 02:59 PM
  4. slopes
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2007, 08:54 AM
  5. Slopes Help
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Nov 14th 2006, 04:27 AM

Search Tags


/mathhelpforum @mathhelpforum