Results 1 to 7 of 7

Math Help - Rotation of a triangle

  1. #1
    Junior Member
    Joined
    Feb 2014
    From
    England
    Posts
    38
    Thanks
    2

    Rotation of a triangle

    A triangle has vertices at the points
    A(6, 9), B(6, 3) and C(9,6).
    Suppose that the triangle is to be moved so that
    B is at the origin
    and
    BC lies along the positive x-axis. One isometry that achieves this
    transformation is the composite of a translation followed by a
    rotation. (You may find it helpful to sketch the triangle.)
    (i) Determine the translation that moves
    B to the origin, giving your
    answer in the form
    ta,b. Write down a formal definition of this
    translation in two-line notation.
    2
    (ii) Find the coordinates of the images
    A! of A and C! of C under the
    translation in part (a)(i).

    (iii) Let
    rθ be the rotation that completes the required
    transformation, where
    θ lies in the interval (π, π]. Find the
    exact values of tan
    θ, cos θ and sin θ, and hence write down a
    formal definition of
    rθ using two-line notation. (There is no need
    to work out the value of the angle
    θ.)

    (iv) Find the coordinates of the images of
    A! and C! under the
    rotation
    rθ. Give your answers as exact values.

    (v) Write down a formal definition of the composite transformation,
    that is, the result of the translation in part (a)(i) followed by the
    rotation in part (a)(iii).


    I've been stuck on this all day! I have t-6,3 for part (i) and found A to be (-12,6) and C to be (3,-9) for part (ii).
    Other than that I'm totally lost!
    Any help would be really appreciated, thanks.






    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,170
    Thanks
    765

    Re: Rotation of a triangle

    Hey alexlbrown59

    Do you have to put your answers in matrix form (where you find the matrices and apply them to the vectors themselves)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2014
    From
    England
    Posts
    38
    Thanks
    2

    Re: Rotation of a triangle

    Hi Chiro, thanks for replying!
    I haven't covered matrix form yet, so no I don't think so.
    Can you still help?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2014
    From
    England
    Posts
    38
    Thanks
    2

    Re: Rotation of a triangle

    Anyone??! Please
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2014
    From
    England
    Posts
    38
    Thanks
    2

    Re: Rotation of a triangle

    I've made slight progress...I have sin theta = 3sqrt10/10, cos theta = sqrt10/10 and tan theta = 3
    Am I on the right lines?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Rotation of a triangle

    Let's work out the translation, first (since we have to APPLY it first). We need to move point B = (6,3) to the origin O = (0,0).

    Since we're going to move "everything at once", let's just subtract 6 from every x-coordinate, and 3 from every y-coordinate. That should do the trick.

    Hence: $$t_{-6,-3}(x,y) = (x-6,y-3)$$

    or:

    $$\begin{array}{l}x' = x - 6\\y' = y - 3 \end{array}$$

    Before we do the rotation, let's think about what we're going to do: we need to put point C on the x-axis.

    Now after the translation, C is at point (9 - 6, -6 - 3) = (3,-9).

    We can form a right triangle with hypotenuse OC. The hypotenuse has length:

    $$\sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$$.

    This tells us:

    $$\cos(-\theta) = \frac{1}{\sqrt{10}},\ \sin(-\theta) = \frac{-3}{\sqrt{10}}$$

    and therefore that:

    $$\cos\theta = \frac{1}{\sqrt{10}},\ \sin\theta = \frac{3}{\sqrt{10}}$$

    Hence:

    $$\tan\theta = 3$$.

    We can write the rotation as:

    $$r_{\theta}(x',y') = (x'\cos\theta - y'\sin\theta,x'\sin\theta + y'\cos\theta)$$

    or:

    $$\begin{array}{l}x'' = x'\cos\theta - y'\sin\theta\\y'' = x'\sin\theta + y'\cos\theta \end{array}$$

    so, explicitly, this becomes:

    $$r_{\theta}(x',y') = \left(\frac{x'- 3y'}{\sqrt{10}},\frac{3x'+y'}{\sqrt{10}}\right)$$

    The composition is thus:

    $$r_{\theta} \circ t_{-6,-3}(x,y) = r_{\theta}(t_{-6,-3}(x,y))$$

    which you should be perfectly capable of evaluating.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2014
    From
    England
    Posts
    38
    Thanks
    2

    Re: Rotation of a triangle

    Thank you so much Deveno!
    I had tan theta = 3, sin theta = 3sqrt10/10 (which is the same as 3/sqrt10) and cos theta = sqrt10/10 (which is the same as 1/sqrt10) so you've proven me right!
    Thanks for taking the time to help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Iso triangle rotation problem
    Posted in the Geometry Forum
    Replies: 0
    Last Post: November 30th 2012, 07:54 AM
  2. Translation and then a rotation of a triangle
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 14th 2010, 07:29 AM
  3. Rotation of a triangle
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 5th 2010, 10:44 AM
  4. Rotation of a triangle
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: February 5th 2010, 05:21 AM
  5. Rotation of triangle
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 31st 2009, 02:20 PM

Search Tags


/mathhelpforum @mathhelpforum