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Thread: 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 $\displaystyle 16 = 6(\sec \theta - \cos \theta)$ for $\displaystyle \cos \theta$ and hence $\displaystyle \theta$.
    Since triangle AEC is isosceles, the angle at E is equal to $\displaystyle (180 - 2 \theta)^o$.
    Now solve for AC using the cosine rule.
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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 $\displaystyle 16 = 6(\sec \theta - \cos \theta)$ for $\displaystyle \cos \theta$ and hence $\displaystyle \theta$.
    Since triangle AEC is isosceles, the angle at E is equal to $\displaystyle (180 - 2 \theta)^o$.
    Now solve for AC using the cosine rule.
    Note: Since you can get an exact value for $\displaystyle \cos \theta$, it's possible to get an exact value for $\displaystyle \cos(180 - 2 \theta)$.
    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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    Hello, geton!


    Draw chord $\displaystyle BC.$

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

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

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

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


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

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

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

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

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

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

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

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


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


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

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