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Math Help - vector please.....

  1. #1
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    Question vector please.....

    If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
    How can I find the value of angle A?
    Please, thank you
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  2. #2
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    Quote Originally Posted by sallsa View Post
    If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
    How can I find the value of angle A?
    Please, thank you
    Hello,

    from the title of your post I assume that you have to use vectors (If you don't want to use vectors you can solve this problem by using Cosine rule)

    1. Calculation of the vectors:

    \overrightarrow{AB}=(1,2)-(3,4)=(-2,-2)

    \overrightarrow{AC}=(6,1)-(3,4)=(3,-3)

    2. The angle \alpha between 2 vectors \vec u and \vec v is calculated by:

    \cos(\alpha)=\frac{\vec u \cdot \vec v}{|\vec u| \cdot |\vec v|}

    Plug in the vectors you know into this formula:

    \cos(A)=\frac{(-2,-2) \cdot (3,-3)}{\sqrt{8} \cdot \sqrt{18}}=\frac{0}{12}=0

    Therefore A = 90 : You are dealing with a right triangle.

    3. Hint: Draw a sketch of this triangle to prove the calculations.

    EB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sallsa View Post
    If I have the triangle ABC, that A(3,4) B(1,2) C(6,1)
    How can I find the value of angle A?
    Please, thank you
    Calculate the squares of the lengths of the sides:

    AB^2=(3-1)^2+(4-2)^2=8

    AC^2=(3-6)^2+(4-1)^2=18

    BC^2=(1-6)^2+(2-1)^2=26.

    So AB^2+AC^2=BC^2. Therefore by the converse of Pythagoras's theorem
    ABC is a right triangle with BC as hypotenuse, and the angle at A is a right
    angle.

    RonL
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