# need help on Slopes

• Nov 18th 2006, 05:21 AM
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 :)
• Nov 18th 2006, 05:39 AM
Plato
Do you have the idea of ‘vectors’ to work with?
We can use the arccosine of dot products to find the angles.
• Nov 18th 2006, 07:13 AM
earboth
Quote:

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
• Nov 18th 2006, 10:05 AM
Soroban

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: . $\tan\theta \:=\:\frac{m_2 - m_1}{1 + m_1\!\cdot\!m_2}$

Quote:

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}
$