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    I'm struggling with this question, I've done part A, I don't know where to start with B.

    https://gyazo.com/20f0a063eab9fa1189aded7966265993
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    Trig - SohCahToa
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    SohCahToa?

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    Help?
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    (Original post by Naruke)
    Help?
    if you rearrange RsinΘ = r ( 1 + sinΘ )

    to make sinΘ the subject it comes out quickly
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    (Original post by the bear)
    if you rearrange RsinΘ = r ( 1 + sinΘ )

    to make sinΘ the subject it comes out quickly
    I did and I got:  Sin \theta = \frac {r+rSin \theta}{R}

    But I don't how understand how they got that, because I've used both sine rule & ratio but I don't get that :/
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    (Original post by Naruke)
    I'm struggling with this question, I've done part A, I don't know where to start with B.

    https://gyazo.com/20f0a063eab9fa1189aded7966265993
    CE=OCsin?
    r=[R-r]sin?
    r+rsin?=Rsin?
    Rsin?=r[1+sin?]

    Sorry, ?=theta
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    (Original post by Naruke)
    I did and I got:  Sin \theta = \frac {r+rSin \theta}{R}

    But I don't how understand how they got that, because I've used both sine rule & ratio but I don't get that :/
    you need to get all the sinΘ together on the same side...
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    (Original post by the bear)
    you need to get all the sinΘ together on the same side...
    I don't understand what you talking about

    but, I've thought of something... would this be ok?

    CE is perpendicular to OB therefore OEC will be  90^o

    As OT bisects O, COE will be  \theta

    so,  Sin \theta = \frac {r}{R-r}

     (R-r)Sin \theta = r

     RSin \theta - rSin \theta = r

     RSin \theta = r + rSin \theta

     RSin \theta = r(1 + Sin \theta)
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    (Original post by Naruke)
    I don't understand what you talking about

    but, I've thought of something... would this be ok?

    CE is perpendicular to OB therefore OEC will be  90^o

    As OT bisects O, COE will be  \theta

    so,  Sin \theta = \frac {r}{R-r}

     (R-r)Sin \theta = r

     RSin \theta - rSin \theta = r

     RSin \theta = r + rSin \theta

     RSin \theta = r(1 + Sin \theta)
    That's fine
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    (Original post by Francesca Meng)
    That's fine
    Thanks
 
 
 
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