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Math Help - Trigonometry

  1. #1
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    Trigonometry

    There is my problem in attachment. I've done (a), but how can I solve (b). Please help me.
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  2. #2
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    Quote Originally Posted by geton View Post
    There is my problem in attachment. I've done (a), but how can I solve (b). Please help me.
    Let the point E be the centre of the circle and consider triangle AEC.
    AE = EC = 3 cm.
    Solve 16 = 6(\sec \theta - \cos \theta) for \cos \theta and hence \theta.
    Since triangle AEC is isosceles, the angle at E is equal to (180 - 2 \theta)^o.
    Now solve for AC using the cosine rule.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Let the point E be the centre of the circle and consider triangle AEC.
    AE = EC = 3 cm.
    Solve 16 = 6(\sec \theta - \cos \theta) for \cos \theta and hence \theta.
    Since triangle AEC is isosceles, the angle at E is equal to (180 - 2 \theta)^o.
    Now solve for AC using the cosine rule.
    Note: Since you can get an exact value for \cos \theta, it's possible to get an exact value for \cos(180 - 2 \theta).
    Do so if you're required to get an exact value for AC ......
    Otherwise use a calculator to get a decimal approximation.
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  4. #4
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    Hello, geton!


    Draw chord BC.

    Since \Delta ACB is inscribed in a semicircle, \angle ACB = 90^o

    In right triangle DBA\!:\;\;\sec\theta \:=\:\frac{AD}{6}\quad\Rightarrow\quad AD \:=\:6\sec\theta

    In right triangle ACB\!:\;\;\cos\theta \:=\:\frac{AC}{6}\quad\Rightarrow\quad AC \:=\:6\cos\theta

    Hence: . CD \;=\;AD - AC \;=\;6\sec\theta - 6\cos\theta


    . . Therefore: . (a)\;\;CD\;=\;6(\sec\theta - \cos\theta)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    In right triangle ACB, we have:
    . . \cos\theta = \frac{AC}{6}\quad\Rightarrow\quad AC \:=\:6\cos\theta .[1]

    Since CD = 16, part (a) gives us: . 6(\sec\theta - \cos\theta) \:=\:16

    Then we have: . 3\left(\frac{1}{\cos\theta} - \cos\theta\right) \:=\:8

    . . which simplifies to: . 3\cos^2\!\theta + 8\cos\theta - 3 \:=\:0

    . . which factors: . (3\cos\theta-1)(\cos\theta +3) \:=\:0

    And we have: . 3\cos\theta - 1 \:=\:0\quad\Rightarrow\quad \cos\theta \:=\:\frac{1}{3} .**


    Substitute into [1]: . AC \:=\:6\left(\frac{1}{3}\right)\quad\Rightarrow\qua  d\boxed{(b)\;AC \:=\:2}


    **
    The other factor produces: . \cos\theta+3 \:=\:0\quad\Rightarrow\quad\cos\theta \:=\:-3
    . . which has no real root.

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