Angle PSLE 2010 P2 Q9

9.  In the figure below, ABCD is a trapezium. E is a point on AD such that AB=BE. Angle BCD=62⁰ and angle CDE=110⁰

Find angle EBC.

180  -  110  =  70

180  -  70  -  70  =  40

180  -  62  =  118

118  -  40  =  78°

It is 78°.

Angle P6 2010 SA2 NY P2 Q8

Angle P6 2010 SA2 NY P2 Q12

P6 Angles A001

/_ ABE = 60 ^o                  (ABE is an equilateral triangle)

/_ EBC = 80 ^o – 60 ^o

                   = 20 ^o

/_ FCB = /_ FBC     (Since the 3 sides of Triangles BCD and

                                  BCF are the same)

            = 20 ^o ÷ 2      (Since EC divides /_ ACB equally
                                   as ABE is an equilateral triangle)

            = 10 ^o

/_ BDC = 180 ^o – 20 ^o – 10 ^o

             = 150 ^o

Angles P6 2009 SA2 P2 MGS Q9

9. In the figure below, not drawn to scale, ABCD is a rhombus. /_BAC = 68^o and /_CED = 45^o.

a) Calculate /_ABC. [1m]

b) Calculate /_CDE. [2m]

180 – 68 – 68 = 44

a) /_ABC = 44^o

68 – 45 = 23

b) /_CDE = 23^o

Angles P6 2009 SA2 P2 Rulang Q15

15. The figure below, not draw to scale, is made up of an isosceles triangle and a rhombus. /_a = /_d, /_c is twice /_b, /_a is twice /_e and /_b is greater than /_e by 48^o.

a) Find the value of /_ e.

b) Find the value of /_ b.

a = d

c = 2 b

a = 2 e

b = e + 48

a + 2 b = 180

2 e + 2 (e + 48) = 180

4 e = 180 – 96

e = 84 ÷ 4 = 21

b = 21 + 48 = 69

a) /_e = 21^o

b) /_b = 69 ^o