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Math Help - Determine if this is a right triangle.

  1. #1
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    Determine if this is a right triangle.

    Given A(5,2), B(-3,6), and C(1,-3), determine if triangle ABC is a right triangle.

    So first I took a look at the angle between A and B:

    cos theta = [5,2][-3,6] / magnitude of [5,2] x magnitude of [-3,6]

    = -15 + 12 / square root of 25+4 x square root of 9+36

    = -3 / square root of 1305

    theta = about 85.2 degrees


    Then I took a look at the angle between B and C:

    cos theta = [-3,6][1,-3] / magnitude of [-3,6] x magnitude of [1,-3]

    = -3 - 18 / square root of 9+36 x square root of 1+9

    = -21 / square root of 450

    theta = about 171.9 degrees

    I didn't even calculate the angle between B and C, and i'm already over 180 degrees so where did I go wrong?
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  2. #2
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    Consider the three distances AB,~AC,~\&~,BC.
    Do they form a Pythagorean Triple?
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  3. #3
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    But the 2 angles I calculated go over 180 degrees (what a triangle is made up of by all of its angles), so where did I go wrong in my calculations?
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  4. #4
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    You did not go wrong. You just find the other angle which is bitween AB and BC, and that is the angle 171.9. The angle that you need is 180-171.9=28.1.

    Just draw the points on coordinate system, and see what angles you need.

    Regards.
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  5. #5
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    But the angle between A and B is 85.2 degrees + the angle between B and C is 171.9 degrees = 257.1 degrees

    Am I missing something? 257.1 degrees is way over 180 degrees Don't I have to add all 3 angles? AB + BC + AC = 180 degrees.. So far I only calculated AB and BC and i'm at 257.1 degrees.
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  6. #6
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    Quote Originally Posted by kmjt View Post
    Given A(5,2), B(-3,6), and C(1,-3), determine if triangle ABC is a right triangle.

    So first I took a look at the angle between A and B:

    cos theta = [5,2][-3,6] / magnitude of [5,2] x magnitude of [-3,6]

    = -15 + 12 / square root of 25+4 x square root of 9+36

    = -3 / square root of 1305

    theta = about 85.2 degrees


    Then I took a look at the angle between B and C:

    cos theta = [-3,6][1,-3] / magnitude of [-3,6] x magnitude of [1,-3]

    = -3 - 18 / square root of 9+36 x square root of 1+9

    = -21 / square root of 450

    theta = about 171.9 degrees

    I didn't even calculate the angle between B and C, and i'm already over 180 degrees so where did I go wrong?

    You should consider the angle between BA and BC and so on. You are calculating the angle whose vertex is at the origin and subtending AB.
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  7. #7
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    you do not calculate the angle between two points ... you calculate the angle between two vectors.

    the angle between vector AB and vector AC is angle A.

    AB = \left<-8,4\right>

    AC = \left<-4,-5\right>

    A = \arccos\left(\frac{AB \cdot AC}{|AB| |AC|}\right)

    A = \arccos\left(\frac{12}{\sqrt{3280}}\right) = 77.9^\circ
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  8. #8
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    Taking Plato's advice, i.e. checking whether the points A, B, and C are the vertices of a right-angled triangle, is probably more neat.
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