RGS Primary 2009 PSLE Math Prelim Paper 2 Q8

In the figure below, O is the centre of the circle where OCD is an equilateral triangle. Given that Angle OAB = 20 degrees and Angle AOD = 127 degrees, find Angle BOC.

Solution

Angle DOC --> 60 degrees (Triangle OCD is equilateral)
Angle AOB --> (180 - 20 - 20) degrees
= 140 degrees (Triangle OAB is isosceles)

Angle BOC --> (360 - 140 - 127 - 60) degrees
= 33 degrees

Answer: 33 degrees

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ACS Primary 2009 PSLE Math Prelim Paper 2 Q13

In the figure below, O is the centre of the circle and AE is parallel to BC. DF = DE, Angle OAB = 58 degrees and Angle FED = 50 degrees.

a) Find Angle GBC
b) Find Angle DCB

Solution

a)
Angle ABO = 58 deg (isosceles triangle)
Angle BOG = (58 + 58) deg = 116 deg (exterior angles)
Angle OGB = [(180 - 116) divided by 2] = 32 deg
Angle GBC = 32 deg (alternate angles)

Answer: 32 degrees


b)
Angle FDE = (180 - 50 - 50) deg = 80 deg
Angle FDE = Angle GDC = 80 deg
Angle DCB = (180 - 80) deg = 100 deg

Answer: 100 degrees
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CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q6

ABE is an isosceles triangle and BCDE is a rhombus. AED is a straight linie. Given that Angle ECG = 16 degrees, find Angle GCD.

Solution

Ange BEA --> (180 - 33) divided by 2 = 73.5
Angle BED --> 180 - 73.5 = 106.5
Angle CED --> 106.5 divided by 2 = 53.25
Angle ECD = Angle CED = 53.5
Angle GCD = 53.25 - 16 = 37.25

Answer: 37.25 degrees

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q10

In the figure shown below, SUVX is a square. STU is an equilateral triangle and TXW is a straight line.
a) Find the value of Angle STX.
b) Find the value of Angle WVX.

Solution

a)
Line TZ passes through V, while line TY is passes through the centre of Line SU.
Angle STX is 1/4 of Angle STU.
Angle STU is 60 degrees (Triangle STU is equilateral)

Angle STX --> (1/4) x 60 degrees = 15 degrees

Answer: 15 degrees

b)
Angle SXT = 15 degrees (Triangle STX is isosceles)
Angle TXV --> (90 - 15) degrees = 75 degrees
Angle WXV --> (180 - 75) degrees = 105 degrees

Angle WVX
--> (180 - 105) degrees divided by 2 = 37.5 degrees
(Triangle WVX is isosceles)

Answer: 37.5 degrees

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Nanyang Pri Sch 2009 P6 CA1 Math Paper 2 Q1

LMNO is a square. PQN and PLQ are isosceles triangles. Angle QNM is 23 degrees. Find Angle NPQ.

Solution

Angle PNO --> 23 deg (mirror image of Angle MNQ)

Angle OPN --> 90 deg - 23 deg = 67 deg

Angle LPQ --> 45 deg (Triangle LQP is isosceles and Angle LPQ is a rt angle)

Angle NPQ --> (180 - 45 - 67) deg = 68 deg

Answer: 68 degrees

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Rosyth Sch 2007 PSLE Math Prelim Q39

Find the sum of the six marked angles in the diagram.

The sum of the all the angles in the triangle in the centre is 180 degrees. The sum of all the 3 angles outside the centre triangle, in the figure above is therefore also 180 degrees (opposite angles).

The sum of all the six marked angles is hence,

Sum of marked angles in 1 triangle ----- 180 degrees
Sum of marked angles in 3 triangles ----- 180 degrees x 3 = 540 degrees
Sum of 6 marked angles ----- (540 – 180) degrees = 360 degrees

Answer: 360 degrees

Anglo Chinese School 2007 PSLE Math Prelim Q38

In the figure not drawn to scale, ABCD is a square. BDE and BCF are equilateral triangles. What is Angle FGE?




Solution

Angle BCF ---- 60º (Triangle BCF is an equilateral triangle)

Angle CBD ---- 45º

Angle CBG ---- 60º - 45º = 15º

Angle BGC ---- 180º - 60º - 15º = 105º (sum of angles in a triangle)

Angle FGE ---- 105º (opposite angles)


Answer: 105º

Rosyth Sch 2006 PSLE Math Prelim Q41

In the figure below, AB is parallel to DE and ACDG is parallel to EF. ABC is an isosceles triangle with Angle ABC = 36 degrees. Find Angle DEF.



Solution

Angle BAC ---- (180 – 36) degrees divided 2 = 72 degrees
Angle EDG ----- 72 degrees
Angle DEF ----- (180 – 72) degrees = 108 degrees (Answer)

Tao Nan School P5 SA2 2006 Math Q44

ABDE is a parallelogram and BCD is an isosceles triangle. Calculate
a) Angle DBC
b) Angle EAB
The figure below is not drawn to scale.



Solution

a) Angle DBC ----- (180 – 42 – 42) degrees = 96 degrees (Answer)

b) Angle EAB = Angle EDB -----
(96 + 42) degrees = 138 degrees (Answer)

Nanyang Pri Sch P6 CA1 2008 Math Q36

The figure below is made up of Triangle FHJ and Trapezium ABDE. Given that GB // FC and BJC is an isosceles triangle, find Angle GHB.



Solution

Angle BCF ----- 106 deg *
Angle BCJ ----- (180 – 106) deg = 74 deg **
Angle BJC ----- 74 deg #
Angle GBH ---- 74 deg *
Angle GHB ----- Angle BGF – Angle GBF ##
= (156 – 74) deg = 82 degrees (Answer)

* corresponding angles
** angles on a straight line
# base angles of isosceles triangle
## exterior angle

Tao Nan School P5 SA2 2007 Math Q44

ACEG is a square, not drawn to scale. Lines DF, CG and BH are parallel lines. Angle EFD is 45 degrees. Calculate
a) Angle CDF
b) Angle AHB



Solution

a)
Angle EDF = (90 – 45) degrees
= 45 degrees

Angle CDF = (180 – 45) degrees
= 135 degrees (Answer)


b)
Angle EFD = Angle AHB = 45 degrees (Answer)

Henry Park Pri Sch 2007 PSLE Math Prelim Q44

Study the diagram below.
a) What is the sum of Angle a, Angle b, Angle c, Angle d, Angle e and Angle f?


Solution

Angles on a straight line = 180 degrees
3 straight lines will have a total of 180 degrees x 3 = 540 degrees
Sum of all Angles a to f -----
540 degrees – sum of angles of triangle
= (540 – 180) degrees
= 360 degrees (Answer)


The figure below shows a rectangle and a triangle.
b) What is the sum of Angle p + Angle q?


Solution

Sum of interior angles of 2 parallel lines ----- 180 degrees (Answer)

Henry Park Pri Sch 2007 PSLE Math Prelim Q39

The figure is not drawn to scale. ABCD is a parallelogram. Find Angle ADC.



Solution

Angle AEB = (180 - 58 – 35) degrees = 87 degrees
(Angles in a triangle)

Angle ABC = (87 + 38) degrees = 125 degrees

Angle ADC = 125 degrees (Answer) (Opposite angles of a parallelogram)

MGS (Paya Lebar) Pri School P6 Math CA1 2006 (Q44)

In the figure not drawn to scale, PQRS is a parallelogram, PX = PY and YXQ is a straight line. Find
(a) Angle PQY
(b) Angle YPS





Solution

(a) Find Angle PQY

Angle QPS + Angle PSR = 180 degrees because PQRS is a parallelogram.

Therefore, Angle SPQ = 180 degrees – 125 degrees = 55 degrees.

Angle PQY = 180 degrees – 95 degrees – Angle SPQ

= (180 – 95 – 55) degrees = 30 degrees

Answer: Angle PQY is 30 degrees.


(b) Find Angle YPS

Angle SPX = Angle SPQ – Angle QPX
= 55 degrees – 32 degrees
= 23 degrees

Angle PXY = 180 degrees – 95 degrees – Angle SPX
= (180 – 95 – 23) degrees
= 62 degrees

Angle PYX is also 62 degrees because triangle PYX is an isosceles triangle.
Angle PZY = (180 – 95) degrees = 85 degrees

Angle YPS = 180 degrees – Angle PZY – Angle PXY
= (180 – 85 – 62) degrees
= 33 degrees

Answer: Angle YPS is 33 degrees.