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    for part i) I have no idea on how to find angle BAC, I was thinking about trying to find the length BC or AC, but im not sure thats what im supposed to do.
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    Part i) It is a root therefore the y valur there is 0. Pluggin lambda in we get:
    2 (2cos (lam)-sin (lam))sin (lam)=0
    Cancelling the and sinlam we get:
    2cos (lam )-sin (lam)=0
    2cos (lam)=sin (lam)

    Dividing by coslam to both sides:
    Sinlam/coslam=2
    Tan(lam)=2.
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    (Original post by Radioactivedecay)
    Part i) It is a root therefore the y valur there is 0. Pluggin lambda in we get:
    2 (2cos (lam)-sin (lam))sin (lam)=0
    Cancelling the and sinlam we get:
    2cos (lam )-sin (lam)=0
    2cos (lam)=sin (lam)

    Dividing by coslam to both sides:
    Sinlam/coslam=2
    Tan(lam)=2.
    Haha thanks mate, Not sure whats happened but I cant seem to get rid of that image, it for a different question.
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    Fixed the OP, now the image corresponds to the actual question
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    (Original post by znx)
    Haha thanks mate, Not sure whats happened but I cant seem to get rid of that image, it for a different question.
    Oh haha ok then. Thats fine.

    for part i of the attachement, just look at the point A. you know that the sum of all angles must equal 360 as it goes around one entire circle, so to speak. So you get
    BAC +120+90-theta+90=360

    then you simply rearrange.
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    (Original post by Radioactivedecay)
    Oh haha ok then. Thats fine.

    for part i of the attachement, just look at the point A. you know that the sum of all angles must equal 360 as it goes around one entire circle, so to speak. So you get
    BAC +120+90-theta+90=360

    then you simply rearrange.
    Thank you mate
 
 
 

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