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

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    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.
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    Quote Originally Posted by help1 View Post
    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
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    Hello, help1!

    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

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