# Points A(-1,4), B(3,12) and C(7,-2) are given. Find (to the nearest tenth of a degre

• April 17th 2008, 04:35 PM
help1
Points A(-1,4), B(3,12) and C(7,-2) are given. Find (to the nearest tenth of a degre
Points A(-1,4), B(3,12) and C(7,-2) are given. Find (to the nearest tenth of a degree) the measure of <BAC.
• April 17th 2008, 05:39 PM
topsquark
Quote:

Originally Posted by help1
Points A(-1,4), B(3,12) and C(7,-2) are given. Find (to the nearest tenth of a degree) the measure of <BAC.

You can do this in two ways:
Vectors: Find vectors BA and BC. Then
$cos(\theta) = \frac{\vec{BA} \cdot \vec{BC}}{|BA| |BC|}$

Triangles
Find sides AB, AC, BC. Then use the Law of cosines:
$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC ~cos(\theta)$

-Dan
• April 17th 2008, 10:23 PM
Soroban
Hello, help1!

Quote:

Points $A(-1,4),\; B(3,12),\;C(7,-2)$ are given.
Find (to the nearest tenth of a degree) the measure of $\angle BAC$

Are you familiar with this formula?

The angle $\theta$ between two lines with slopes $m_1\text{ and }m_2$

. . is given by: . $\tan\theta \;=\;\frac{m_2 - m_1}{1+m_1m_2}$

We have: . $m_1\:=\:m_{AC} \;=\;\frac{\text{-}2-4}{7-(\text{-}1)} \;=\;-\frac{3}{4}$

. . . . and: . $m_2 \:=\:m_{AB} \:=\:\frac{12-4}{3-(\text{-}1)} \:=\:2$

Hence: . $\tan\theta \;=\;\frac{2 - \left(\text{-}\frac{3}{4}\right)}{1 + (2)\left(\text{-}\frac{3}{4}\right)} \;=\;-\frac{11}{2}$

Therefore: . $\theta \;=\;\tan^{-1}\!\left(\text{-}\frac{11}{2}\right) \;\approx\;100.3^o$