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Math Help - circle and triangle trig help

  1. #1
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    circle and triangle trig help

    In the diagram AB=6 cm is the diameter of the circle and BT is the tangent to the circle at B. The chord AC is extended to meet this tangent at D and ∠DAB=θ.
    (a) Show that CD=6(secθ−cosθ) .
    (b) Given that CD=16 cm, calculate the length of the chord AC.
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  2. #2
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    Quote Originally Posted by Tweety View Post
    In the diagram AB=6 cm is the diameter of the circle and BT is the tangent to the circle at B. The chord AC is extended to meet this tangent at D and ∠DAB=θ.
    (a) Show that CD=6(secθ−cosθ) .
    (b) Given that CD=16 cm, calculate the length of the chord AC.

    (a) first , we find AD .

    cos\theta=\frac{6}{AD}

    AD=6sec\theta

    Then now AC .

    sin(90-\theta)=\frac{AC}{6}

    AC=6cos\theta

    AD-AC=CD

     <br />
CD=6sec\theta-6cos\theta<br />

     <br />
CD=6(sec\theta-cos\theta)<br />
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  3. #3
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    Quote Originally Posted by Tweety View Post
    In the diagram AB=6 cm is the diameter of the circle and BT is the tangent to the circle at B. The chord AC is extended to meet this tangent at D and ∠DAB=θ.
    (a) Show that CD=6(secθ−cosθ) .
    (b) Given that CD=16 cm, calculate the length of the chord AC.
    (a) The key to this is the 'angle in a semicircle is a right-angle' property; i.e. \angle ACB = 90^o. And, of course, the angle between a radius (or a diameter) and a tangent is also a right-angle. So \angle ABD = 90^o

    So in \triangle ABC, AC = 6 \cos \theta

    and in \triangle ABD, AD = 6\sec\theta

    \Rightarrow CD = 6(\sec\theta - \cos\theta)

    For part (b): 16 = 6(\sec\theta - \cos\theta)

    \Rightarrow 3\sec\theta - 3\cos\theta = 8

    \Rightarrow 3 - 3\cos^2\theta = 8\cos\theta

    \Rightarrow 3\cos^2\theta + 8\cos\theta - 3 = 0

    \Rightarrow (3\cos\theta - 1)(\cos\theta + 3)=0

    \Rightarrow \cos\theta = \tfrac13 ( -3 being impossible)

    \Rightarrow AC =6\cos\theta = 2

    Grandad
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    (a) first , we find AD .

    cos\theta=\frac{6}{AD}

    AD=6sec\theta

    Then now AC .

    sin(90-\theta)=\frac{AC}{6}

    AC=6cos\theta

    AD-AC=CD

     <br />
CD=6sec\theta-6cos\theta<br />

     <br />
CD=6(sec\theta-cos\theta)<br />
    thanks,
    but I can't seem to follow from this step
    sin(90-\theta)=\frac{AC}{6}

    how did you get this?
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  5. #5
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    Quote Originally Posted by Tweety View Post
    thanks,
    but I can't seem to follow from this step
    sin(90-\theta)=\frac{AC}{6}

    how did you get this?
    I assume that it's from the right triangle ABC. Angle ABC would be 90^{\circ} - \theta.


    01
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  6. #6
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    Quote Originally Posted by Tweety View Post
    thanks,
    but I can't seem to follow from this step
    sin(90-\theta)=\frac{AC}{6}

    how did you get this?

    erm ok , from your diagram , you draw another line from C to B and angle ACB is 90 degrees so angle ABC is 90-theta
    Then you will get :

    sin(90-\theta)=\frac{AC}{6}
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    erm ok , from your diagram , you draw another line from C to B and angle ACB is 90 degrees so angle ABC is 90-theta
    Then you will get :

    sin(90-\theta)=\frac{AC}{6}
    why is it sin\theta ? should it not be cos\theta ?

    thats what i dont get;

    Thanks.
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  8. #8
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    Quote Originally Posted by Tweety View Post
    why is it sin\theta ? should it not be cos\theta ?

    thats what i dont get;

    Thanks.

     <br />
sin(90-\theta)=cos\theta<br />
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