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    hi, i just came across a question and i was wondering if my answer is correct.
    so i attained an answer of 119.5196 degrees and was wondering if it was correct because i think it's rather odd.

    thank you
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    (Original post by User23)
    hi, i just came across a question and i was wondering if my answer is correct.
    so i attained an answer of 119.5196 degrees and was wondering if it was correct because i think it's rather odd.

    thank you
    What was your method?
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    (Original post by 16Characters....)
    What was your method?
    So basically using the cosine rule i found out the length of BC which gave me 6.89258cm. then i reaaranged the cosine rule to make CosB the subject and then used length BC to find it out. is this correct.. or?
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    I got 62.3

    EDIT: Nvm, my calculator was in radians
    But now I get 60.5
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    (Original post by User23)
    So basically using the cosine rule i found out the length of BC which gave me 6.89258cm. then i reaaranged the cosine rule to make CosB the subject and then used length BC to find it out. is this correct.. or?
    After Cos rule to get the other length, you do sine rule: Sin(x)/12 = Sin(30)/6.89...

    Also, never round off answers during the working out, only round off the final answer to the specified amount of d.p's

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    (Original post by eternaforest)
    I got 62.3

    EDIT: Nvm, my calculator was in radians
    But now I get 60.5
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    I suspect you used the sine rule, which doesn't distinguish between angles less than 90 and those greater than 90.
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    (Original post by User23)
    hi, i just came across a question and i was wondering if my answer is correct.
    so i attained an answer of 119.5196 degrees and was wondering if it was correct because i think it's rather odd.

    thank you
    I got 119.48... without using any truncation in the calculation.
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    (Original post by User23)
    So basically using the cosine rule i found out the length of BC which gave me 6.89258cm. then i reaaranged the cosine rule to make CosB the subject and then used length BC to find it out. is this correct.. or?
    Yep sounds good.
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    (Original post by ghostwalker)
    I suspect you used the sine rule, which doesn't distinguish between angles less than 90 and those greater than 90.
    Oh right damn, could you use sine rule to find the other angle first?


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    (Original post by User23)
    So basically using the cosine rule i found out the length of BC which gave me 6.89258cm. then i reaaranged the cosine rule to make CosB the subject and then used length BC to find it out. is this correct.. or?
    Your way's much simpler than mine. I got 119.4831..., so be careful with your rounding.
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    (Original post by eternaforest)
    Oh right damn, could you use sine rule to find the other angle first?


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    I did it this way. I needed to use double angle formula to get the answer from there though.
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    (Original post by morgan8002)
    Your way's much simpler than mine. I got 119.4831..., so be careful with your rounding.
    what was ur way?
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    (Original post by User23)
    what was ur way?
    SIne rule.
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    (Original post by morgan8002)
    SIne rule.
    but i dont think u can do that with only one known angle?
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    (Original post by User23)
    but i dont think u can do that with only one known angle?
    I would use the cosine rule to find the missing side and then you can use the sine rule to find the angle


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    (Original post by User23)
    but i dont think u can do that with only one known angle?
    You know the angles of a triangle add up to 180^o, so the last angle is 150-x.
    Applying sine rule we get \frac{\sin x}{12} = \frac{\sin(150-x)}{7}.
    Then some algebra. If you haven't done double angle formulae I don't think you can do it that way though.
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    (Original post by User23)
    but i dont think u can do that with only one known angle?
    You can use sine rule to find the angle that's not x because like ghostwalker said it doesn't distinguish between acute and obtuse angles so finding the acute angle first then subtract that and 30 from 180 (i think)

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    (Original post by fluzzles98)
    I would use the cosine rule to find the missing side and then you can use the sine rule to find the angle


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    The issue with that method is that it leaves you with two possible values of x, x = 60.5 and x = 180-60.5 = 119.5. You would then have to check which angle corresponds to the correct side lengths, which is extra work.
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    (Original post by morgan8002)
    You know the angles of a triangle add up to 180^o, so the last angle is 150-x.
    Applying sine rule we get \frac{\sin x}{12} = \frac{\sin(150-x)}{7}.
    Then some algebra. If you haven't done double angle formulae I don't think you can do it that way though.

    damn ive never heard of double angle formulae before, r u in sixth form or something?
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    (Original post by User23)
    damn ive never heard of double angle formulae before, r u in sixth form or something?
    Don't worry then, use cosine rule repeatedly.

    No, uni. You learn double angle formulae at A-level though.
 
 
 
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