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# helppppppp, difficult maths question watch

1. 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.
2. would be helpful with a picture of the triangle mabye
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.
4. (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
Attached Images

5. Draw a pic then use pithagourous theory ( a^2+b^2=c^2)
6. Is it 32 ?

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7. (Original post by PlayerBB)
Is it 32 ?

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I'm not sure. How did you get 32?
8. 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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9. (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
10. I got 6 root 38
(rushed)
11. (Original post by Scott_dp)
I got 6 root 38
(rushed)
12. 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
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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14. (Original post by PlayerBB)

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Okay
15. 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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16. (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
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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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
Attached Images

19. (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.
20. (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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