06 HSC paper (1 Viewer)

[plak]

Having troubles trying to find the reasons and y >< for 06 HSC paper 10bii.

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_06.pdf

 

[Ali92l]

Let Angle MLN = x ;
Angle KLQ = 90 - x (supplementary Angles )

in Triangle LMN , Angle LMN = 90- x ( Angle sum of Triangle )
in triangle QKL, Angle QKL = x ( Angle sum of trangle )
....

Angle MLN = Angle QKL ( x )
Angle LMN = Angle KLQ ( 90 - x )
Angle MNL = Angle KQL ( right angles given )

Hence.. triangles are similar as they are equiangular

To get y apply the fact that sides in similar triangles are in the same ratios

y / MN = KL / QL

We know KL = 6 + x and QL = root 24x = 2root(6x) and MN = 12

y / 12 = 6+ x / 2root (6x)

y = 12(6+x) / 2root(6x) ( 12/2 comes down to 6/1 ; root(6x) = root 6 x root x )

y = 6/ (6+ x ) / root(6) x root(x)

y = root 6 (6+x) / root(x) (as 6 / root6) is root6 )

 

[plak]

Ahh thanks