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    Since theta is obtuse, the triangle is an ambiguous case and therefore, I can't use trig because when I rearrange for sin C I get 1.12 and not between 0 and 1.

    So how would I solve this... Somehow I imagine multiplying something by 2.
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    I would imagine you use  A =\displaystyle\frac{1}{2}r^{2} \theta

    Unless I am wrong, feel free to correct me anyone!
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    (Original post by boromir9111)
    I would imagine you use  A =\displaystyle\frac{1}{2}r^{2} \theta

    Unless I am wrong, feel free to correct me anyone!
    Still I get theta = 0.64

    ???
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    (Original post by I'm clever)
    Still I get theta = 0.64

    ???
    What's the answer supposed to be?
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    (Original post by boromir9111)
    What's the answer supposed to be?
    If I knew that why would I be asking the question... -.-
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    (Original post by I'm clever)
    If I knew that why would I be asking the question... -.-
    Then my method is correct!
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    (Original post by boromir9111)
    Then my method is correct!
    Ok I think I have it:

    Since theta is obtuse I CAN use the Sin rule. I'll have to do 180 - theta.

    A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta

    \theta = 0.64

     \theta = sin^{-1} (0.64) =39.79

    180-39.79 = 140
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    (Original post by I'm clever)
    Ok I think I have it:

    Since theta is obtuse I CAN use the Sin rule. I'll have to do 180 - theta.

    A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta

    \theta = 0.64

     \theta = sin^{-1} (0.64) =39.79

    180-39.79 = 140
    I didn't even read the question but you should be working in radians.
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    (Original post by I'm clever)
    A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta
    That's the formula for the area of sector OAB rather than the triangle OAB. You don't know what the area of the sector is so you can't deduce that step above.

    What I would do is use formula for the area of the triangle (A=\frac{1}{2}ab\sin C) to find that \frac{1}{2}5^2 \sin \theta = 8 and just solve for \sin \theta. Then recall that \sin \theta = \sin (\pi - \theta).
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    (Original post by boromir9111)
    I would imagine you use  A =\displaystyle\frac{1}{2}r^{2} \theta

    Unless I am wrong, feel free to correct me anyone!
    And the area of the triangle is \displaystyle\frac{1}{2}r^{2} \sin \theta
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    (Original post by Mr M)
    And the area of the triangle is \displaystyle\frac{1}{2}r^{2} \sin \theta
    A=\frac{1}{2}absinC

    8=\frac{1}{2}(5^2)sinC

    sinC=39.79

    Since theta is obtuse 180-39.79=140=2.44 rad.

    Is that correct?
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    (Original post by I'm clever)
    A=\frac{1}{2}absinC

    8=\frac{1}{2}(5^2)sinC

    sinC=39.79

    Since theta is obtuse 180-39.79=140

    Is that correct?
    Wait never mind thats good.
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    (Original post by I'm clever)
    A=\frac{1}{2}absinC

    8=\frac{1}{2}(5^2)sinC

    sinC=39.79

    Since theta is obtuse 180-39.79=140

    Is that correct?
    I hope you don't mind if I shout:

    RADIANS
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    (Original post by Mr M)
    I hope you don't mind if I shout:

    RADIANS
    EDIT: I updated my post before you posted this.
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    (Original post by I'm clever)
    A=\frac{1}{2}absinC

    8=\frac{1}{2}(5^2)sinC

    sinC=39.79

    Since theta is obtuse 180-39.79=140

    Is that correct?
    No, that's the value of \sin \theta for this angle theta and think about it; if your angle is 140 radians (roughly 44.5 pi, which equates to roughly 8010 degrees!), then that's a little too big isn't it? So that's a tell-tale sign that you've made a mistake somewhere. :p:

    You need to inverse sine both sides (using you calculator) and then, if you get an acute angle for theta, consider \pi - \theta so that it will be the acute version. I've explained why this works in my first post.
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    (Original post by I'm clever)
    EDIT: I updated my post before you posted this.
    Well it is still wrong to 3 sig figs which is the required accuracy.

    Your calculator has a radians mode. Use it.
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    (Original post by Mr M)
    Well it is still wrong to 3 sig figs which is the required accuracy.

    Your calculator has a radians mode. Use it.
    sin^{-1}(0.64)=0.694 rad => \theta = \pi-0.694=2.45rad

    Correct now?
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    (Original post by I'm clever)
    sin^{-1}(0.64)=0.694 rad => \theta = \pi-0.694=2.45rad

    Correct now?
    Yes. Try to get used to working in radians as you will only occasionally use degrees from now on. Don't round answers prematurely - write down plenty of figures for intermediate steps and only round the final answer.
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    (Original post by Mr M)
    Yes. Try to get used to working in radians as you will only occasionally use degrees from now on. Don't round answers prematurely - write down plenty of figures for intermediate steps and only round the final answer.
    Well I learnt how to store my answer in the calculator in Statistics so I can apply this to C2, or any question that requires multiple stages of working in maths.
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    But aren't you clever 'I'm clever'?
 
 
 
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