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Q6 (b) (ii) (1 Viewer)

gerardk

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basic run through on how to do it please.

Thanks

(the 2nd part of it)
 

Thomas_wow

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<BAY = < YAM .'. = < MCY

.'. AYB = <AYM = <MYC - all on straight line

.'. 180 = 3x

x = 60

.'. MCY = 30 (180 - 60 - 90)

ratio

YM : CM = 1:2

.'. MY : AC = 1:4

thats what i did anyway ;]... prolly completely wrong tho ;[...
 

quarxer

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let <YAM = x
therfore <BAC = 2x
therfore <BCA = 90 - 2x (angle sum of triangle)
therfore 90 - 2x = x (congruent traingles)
x = 30
<BCA = 90 - 2(30)
=30

Is how i did it, hope its right
 

acmilan

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i got YCM = 30 degrees

then in triangle AMY, tan30 = MY/AM
.'. AM = root3MY

AC = 2AM
.'. AC = 2root3MY
AC/MY = 2root3
MY/AC = 1/2root3
 

gerardk

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quarxer - the second part asks for the ratio MY : AC - thats the part i need help with

ilovechildren - t results is 3U yeah? are you allowed to use 3U methods in 2U?

edit: hey yeah 1 : 2(root)3 is what i got too..if thats the right answer 3marks is alot to give for such a little question
 

acmilan

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I dont know whether you can or cant but they probably would have to give you atleast part marks id think
 

blackmagic219

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yeah i got the same as gerark and acmilan 1987.....few it thought that was completely wrong...
 

Willmaker

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I used the sine rule.

Let YM=x and MC = y

x/sin30 = y/sin60 but since there are 2 lots of y
x/sin30 = 2y/2sin60

Therefore x/2y = ratio = sin30/2sin60

0.5/2(root 3)

But im probably wrong...... >.<
 

Sweet_Lemon

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hum...think i got 1: 2(where's the root thino on this computer~~dam)3.....

should be rite then~~ let MY be 1 i think~~ makes life easier~~ ekekek
 

gerardk

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so why 3 marks for it - i guess 1 mark for the correct angle, but for just saying its a 30 60 triangle and then sayin 1 : 2(root)3 isnt worth 2 marks..maybe they expect u to use the sine rule or whatever
 

bevstarrunner

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I got the angle, then i identified that it was the 1:2:(root)3 triangle then went from there...I dont think they can mark you down for that...
 

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