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 By: parent603 (offline)  Sunday, July 13 2014 @ 05:11 AM CDT (Read 2300 times)
parent603

Hi,

Appreciate if you could help to provide solutions to Q25 and Q26.

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: 625

 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: 625

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