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    Can any one help me with this maths problem? Any help would be useful, thanks

    AC=16cm
    Angle ABC=90 degrees
    Angle CAB= 30 degrees

    BC=BD
    CD=12cm

    Calculate the area of triangle BCD. Give your answer correct to 3 significant figures.
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    would be helpful with a picture of the triangle mabye
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    (Original post by AceOfSpade)
    Can any one help me with this maths problem? Any help would be useful, thanks

    AC=16cm
    Angle ABC=90 degrees
    Angle CAB= 30 degrees

    BC=BD
    CD=12cm

    Calculate the area of triangle BCD. Give your answer correct to 3 significant figures.
    Maybe a sketch helps you to understand? so you (and we) can see the coherences of that construction.

    From the clues you have given, I would advice you to begin with the triangle ABC and to calculate BC and AB.
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    (Original post by Kallisto)
    Maybe a sketch helps you to understand? so you (and we) can see the coherences of that construction.

    From the clues you have given, I would advice you to begin with the triangle ABC and to calculate BC and AB.
    I hope this helps
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    Draw a pic then use pithagourous theory ( a^2+b^2=c^2)
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    Is it 32 ?

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    (Original post by PlayerBB)
    Is it 32 ?

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    I'm not sure. How did you get 32?
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    h= 16Sin30/Sin90
    h=8 then area of triangle BCD= 0.5*8*12*Sin(45) = 33.9
    P.s sorry wrong calculation, not 32 but 33.9

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    (Original post by PlayerBB)
    h= 16Sin30/Sin90
    h=8 then area of triangle BCD= 0.5*8*12*Sin(45) = 33.9
    P.s sorry wrong calculation, not 32 but 33.9

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    Thank you so much, i'm not sure if that's the answer because I don't have the answers but thanks anyway
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    I got 6 root 38
    (rushed)
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    (Original post by Scott_dp)
    I got 6 root 38
    (rushed)
    How, please?
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    CB=16sin(30)=8
    BC=BD=8
    So triangle CBD is isoceles with side lengths 12, 8 and 8 cm.
    Use pythagoras to find height:
    Root(8^2-6^2)=Root(38)
    Area = half base * height = (12/2)*root(38)=6 root 38
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    (Original post by AceOfSpade)
    Thank you so much, i'm not sure if that's the answer because I don't have the answers but thanks anyway
    You're welcome and when you have the answer, post it please

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    (Original post by PlayerBB)
    You're welcome and when you have the answer, post it please

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    Okay
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    But from the pic you have submitted, it seems that ACD is 90 degrees, if it is the BCD is 30 and therefore since BC=BD the BC= 8 from Pythagoras theorm and sin30*8*12*0.5 = 24 so the area is 24 if the angle is as the pic which is 90

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    (Original post by PlayerBB)
    But from the pic you have submitted, it seems that ACD is 90 degrees, if it is the BCD is 30 and therefore since BC=BD the BC= 8 from Pythagoras theorm and sin30*8*12*0.5 = 24 so the area is 24 if the angle is as the pic which is 90

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    They haven't given me the angle for ACD so it does look like 90degrees but I'm not 100% sure
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    (Original post by AceOfSpade)
    They haven't given me the angle for ACD so it does look like 90degrees but I'm not 100% sure
    Hope it is!

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    (Original post by AceOfSpade)
    Can any one help me with this maths problem? Any help would be useful, thanks

    AC=16cm
    Angle ABC=90 degrees
    Angle CAB= 30 degrees

    BC=BD
    CD=12cm

    Calculate the area of triangle BCD. Give your answer correct to 3 significant figures.

    I did this really quickly, based on your diagram, check through to make sure the maths is ok
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    (Original post by rede121)
    I did this really quickly, based on your diagram, check through to make sure the maths is ok
    Your maths looks fine to me but I still don't what the answer is because everyone seems to be getting different ones. But thank you so much for your solution and I will post the answer on Wednesday so you can see if you were correct.
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    (Original post by rede121)
    I did this really quickly, based on your diagram, check through to make sure the maths is ok
    Is the height of the triangle ABC not 8 as 16sin(30)=8 ? which would result in the height of the other triangle being 2root7
 
 
 
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