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Past paper 2 2007 question 7 - HSN forum

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Past paper 2 2007 question 7


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#1 Alex Kirk

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Posted 16 May 2008 - 02:39 PM

Hey

Does anyone know how to do this. I don't know where to start.

Triangle ABC is right-angled at A and BD is the bisector of angle ABC

AB=6 units and CB=10 units

Determine the eaxact value of BD, expressing your answer in the simplest form.

Thanks

Alex


#2 neep

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Posted 16 May 2008 - 04:51 PM

u can work it out using sin rule from standard grade but it dosen't give u an exact value so i think it is wrong?

#3 campbellk2

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Posted 17 May 2008 - 08:55 AM

QUOTE(neep @ May 16 2008, 05:51 PM) View Post
u can work it out using sin rule from standard grade but it dosen't give u an exact value so i think it is wrong?



I was flummoxed my this question too - I worked out the side AC using pythagoras, then halfed it to give myself AD. I know this value is wrong but I didn't know how to find AD. Then I managed to work through to get a wrong value of AB. Hopefully, I would get some followthrough marks, but if anyone could explain how to do it, that would be great!

#4 spoony2oo8

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Posted 17 May 2008 - 02:24 PM

That Question Is Actually Very Difficult.

I Can Actually Not See A Way Of Answeing It.laugh.gif
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#5 spoony2oo8

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Posted 17 May 2008 - 02:53 PM

I Have Managed To Get The Right Asnswer From The Answer Book In The Past Papers But I'm Not Sure If It Would Gain Fulls Marks As Its An Odd Way Of Doing It.

So My First Step Was To Say That : cos2x= 6/10
2x= 53.13
x= 26.56

So Then I Took The Smaller Triangle ABD And Worked Out That The Missing Angle At D Was 63.44 By Doing Angle = 180-90-26.56

Then I Used The Sin Rule To Work Out That BD/sin90 = 6/sin63.44
BD = 6.71
Then I Squared BD To See What It Would Give Me And It Gave Me Almost 45 So I Took That To Be My Answer

BDČ = 45
BD = √45 = √9*√5 = 3√5


And Thats What The Answer Booklet States. Not Sure If Its The Correct Method Though. dry.gif unsure.gif

Anyone Else Figured It Out?
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#6 campbellk2

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Posted 17 May 2008 - 03:29 PM

QUOTE(spoony2oo8 @ May 17 2008, 03:53 PM) View Post
I Have Managed To Get The Right Asnswer From The Answer Book In The Past Papers But I'm Not Sure If It Would Gain Fulls Marks As Its An Odd Way Of Doing It.

So My First Step Was To Say That : cos2x= 6/10
2x= 53.13
x= 26.56

So Then I Took The Smaller Triangle ABD And Worked Out That The Missing Angle At D Was 63.44 By Doing Angle = 180-90-26.56

Then I Used The Sin Rule To Work Out That BD/sin90 = 6/sin63.44
BD = 6.71
Then I Squared BD To See What It Would Give Me And It Gave Me Almost 45 So I Took That To Be My Answer

BDČ = 45
BD = √45 = √9*√5 = 3√5


And Thats What The Answer Booklet States. Not Sure If Its The Correct Method Though. dry.gif unsure.gif

Anyone Else Figured It Out?


Well, I think it's as good a way as any...Thanks for sharing!


#7 Heather-xXx

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Posted 17 May 2008 - 04:36 PM

It took me while to work out too at first but I think this is how you would do it:

First I saw from the diagram that angle ABC is 2x so cos(2x) = 6/10 = 3/5

Then I used the double angle formula to work out cosx:
cos(2x) = 2cos2x - 1 = 3/5
2cos2x = 3/5 + 1 = 8/5
cos2x = 8/5 x 1/2 = 8/10 = 4/5
cosx = √(4/5) = 2/√5

Then I used the standard grade formula thing that says cosx = adjacent/hypoteneuse so...
cosx = 6/BD
2/√5 (from above) = 6/BD
BD = 6/(2/√5)
BD = (6√5)/2
BD = 3√5

Hope this helps.biggrin.gif

#8 spoony2oo8

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Posted 17 May 2008 - 04:50 PM

^That Probably Seems Like The Right Way.

My Answer Was Made On The Assumption That I Had To Square The 6.71 Or Whatever It Was laugh.gif

Everytime I Did The Expantion For cos2x I Ended Up With 4/√5 And Then It Told Me Maths Error So I Think I Was Along The Right Lines The First Time Round laugh.gif

Which Gives Hope For The Exam On Tuesday!
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