
By: parent603 (offline) Sunday, July 13 2014 @ 05:11 AM CDT (Read 2340 times)



parent603 
Hi,
Appreciate if you could help to provide solutions to Q25 and Q26.
Thanks for your help!

Newbie
Registered: 11/07/11 Posts: 14





By: echeewh (offline) Sunday, July 13 2014 @ 01:32 PM CDT



echeewh 
Hey < parent603 >,
Following is my worked solution for Q26.
angle EFC = 180  50 = 130 { sum of internal angles between // lines }
given CBEF is a rhombus, BF is a diagonal that bisects angle EFC.
Hence, angle BFE = angle BFC = 130 ÷ 2 = 65
angle BCF = angle BEF = 50 { diagonally opposite angles are equal }
Given ABCD is a square, angle BCD = 90.
Hence, angle DCF = 90 + 50 = 140
Given triangle DCF is isosceles { CD = CF },
angle CFD = angle CDF = (180  140) ÷ 2 = 20
Hence, angle BFG = 65  20 = 45°
========
Trust this helps.
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627





By: echeewh (offline) Monday, July 14 2014 @ 02:16 AM CDT



echeewh 
Hey < parent603 >,
Kindly refer to attachment for my worked solution to q25.
Trust this helps.
Do let me know again if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627



