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Cindy is some distance NW from a building Don is some distance due North of a building. If Don is twice as far as Cindy from the building and they are 40m apart, calculate their distances to the building to the nearest one-tenth of metre. Can someone please explain step by step on how to do this, I'm like so lost

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Okay. |dw:1316163703925:dw|
Do i calculate x by soh cah toa thing ?
no, the triangle is not guaranteed to be right angled.

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Other answers:

do you know of the sine rule?
Yup, do i use the sine rule ?
use law of cosines or law of sines here to find the angles from the lengths
you can use cosine rule a^2 = b^2+c^2-2bc cos A for angle B
you'll get x that way
Okay, like 40/sin45 = x/sinC ?
w8 nvm
40^2 = x^2 + (2x)^2 - 2x(2x) cos 45 1600 = x^2 + 4x^2 - 4x^2 (1/sqrt(2)) 1600 = 5x^2 - 2sqrt(2)x^2 1600 / (5 - 2sqrt(2)) = x^2 x^2 = 736.79 x = sqrt(736.79) = 27.14m So Don is 27.1*2 = 54.2m away from the building and Cindy is 27.1m away from the building.
Tyvm (:
btw in the diagram I swapped Cindy and Don, but that's not too much of a problem... the problem works itself out in the same way.
Okay xD
Um , I did it and I got x= 47.568 xD
I still you should still get to the same answer...?
idk, thats how i did it 1600 = x^2 + 4x^2 - 4x^2 * cos45 1600 = x^2 *cos45 cos45x^2 = 1600 x^2 = 2262.74 x = 47.578 Did I stuff it up somewhere ?
yes. You first perform multiplication, then subtraction, according to BODMAS. You first applied subtraction and then multiplication!
Hm, i don't follow the BODMAS thing xD Wonder why I stuffed it up xD but how did I stuff it up ? Since I already subtracted it first
1600 = x^2 + 4x^2 - 4x^2 * cos 45 You first performed addition and subtraction: 1600 = x^2 * cos 45 when you should have first performed multiplication! 1600 = x^2 + 4x^2 - 2.828x^2
Where did you get 2.828 from ?
4 * cos 45
Oh yeah my bad i forgot about that Thank you (:

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