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
Stacey Warren - Expert brainly.com
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Do i calculate x by soh cah toa thing ?
no, the triangle is not guaranteed to be right angled.
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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 ?
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.
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.
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