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
$\displaystyle cos(\theta) = \frac{\vec{BA} \cdot \vec{BC}}{|BA| |BC|}$
Triangles
Find sides AB, AC, BC. Then use the Law of cosines:
$\displaystyle AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC ~cos(\theta)$
-Dan
Hello, help1!
Are you familiar with this formula?Points $\displaystyle A(-1,4),\; B(3,12),\;C(7,-2)$ are given.
Find (to the nearest tenth of a degree) the measure of $\displaystyle \angle BAC$
The angle $\displaystyle \theta$ between two lines with slopes $\displaystyle m_1\text{ and }m_2$
. . is given by: . $\displaystyle \tan\theta \;=\;\frac{m_2 - m_1}{1+m_1m_2}$
We have: .$\displaystyle m_1\:=\:m_{AC} \;=\;\frac{\text{-}2-4}{7-(\text{-}1)} \;=\;-\frac{3}{4}$
. . . . and: .$\displaystyle m_2 \:=\:m_{AB} \:=\:\frac{12-4}{3-(\text{-}1)} \:=\:2$
Hence: .$\displaystyle \tan\theta \;=\;\frac{2 - \left(\text{-}\frac{3}{4}\right)}{1 + (2)\left(\text{-}\frac{3}{4}\right)} \;=\;-\frac{11}{2}$
Therefore: .$\displaystyle \theta \;=\;\tan^{-1}\!\left(\text{-}\frac{11}{2}\right) \;\approx\;100.3^o$