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Amy is 20m S30°W from a tree. Belinda is 50m 110° T from the same tree. Find the distance between them, to the nearest metre.

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what do S30°W and 110° T mean ?
Um, I'm not sure but i think South 30 degrees West , 110 degrees true bearing idk. I haven't done it in ages so I forgot xD

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Other answers:

like that ^^^ ?
Um, idk , it looks right though xDD
How can I find the distance between them ?
Um, I don't think that it requires sin and cos, its bearings
What? There's not values of x and y, how can you use the distance formula ?
I got the values of x and y using the sin() and cos() in the table I drew above. Rounded to the nearest metre, I get 64 meters distance.
Okay, tyvm (:
does it make sense ? did you learn this stuff in the past ?
Not in the past xD, I learnt it year 7 and now I forgot xD But I remember that there's an easier way that doesn't require sin and cos
Well , if there is an easier way, I hope someone will post it here - coz I'd like to know it too :)
Hey, you know your diagram what is the angle degrees thing for AB
what do you mean ?
yes we can calculate that - it's 240-110 = 130 degrees
and you're right that we can make the problem simpler
Yep, there's a formula to calculate it then, was it the sine or cosine formula ? or was it soh cah toa ?
I would need to calculate to find out - why are you asking ?
If that angle can be found, then I can use the sine rule xD
are you trying to use the law of the cosines ?
No, the Sine Rule it says it in my book xD
which sine rule ? a/sin(A)=b/sin(B)=c/sin(C) ?
Yes, that's it @fiddle
Yep that xD
yes, but we only have one angle , that 130 degrees at this point.
I know , but the other angle is 30 right ? the one that I drew
why did you pick 30 degrees ?
is it becasue sin(30)=1/2 ?
Well, the question S30°W so that angle is 30 right ?
No, you can't say that...@mimi|dw:1315634688202:dw|
The 30 degrees would be relative to the black line I drew.
yeah, upsilon says the same thing.
Can it be 60 then, since 90-30 = 60 ?
You can't decide like that you need the angle to be in ratio with the opposite sides...
but the law of the cosines might work here.
Yea, that will work
\[d = \sqrt{ a^2 + b^2 - 2*a*b*cos\alpha}\]
Okay, tyvm (:^2%2B50^2-2*20*50*cos%28130%29%29 It's 64 , same answer I got with the first method :)
Ok, ty (:
and it's much shorter than the first method :)
yup xD. Your method looked so complicated before (:

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