# Math Help - need help on Slopes

1. ## 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

2. Do you have the idea of ‘vectors’ to work with?
We can use the arccosine of dot products to find the angles.

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

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

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