The blog postings are about the Singapore Math. The readers can learn from the postings on Solving Singapore Primary School Mathematics. The blog presents the Math Concept, Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can learn the skills to deal with Math Modeling, Math Problem Solving and Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on Angle and Shape.

Read the posting on **Sarin learns Lines and Angles (Math concept)** to understand the concept of lines and angles.

Read the posting on **Sarin learns Shapes and Solids in school (Math Concept)** to know about shapes.

*Challenge yourself with the question before look for the given solution‼!*

**Upper primary school mathematics question UPQ386**

In the figure below, ABCD is a square. CDF is an equilateral triangle. AF is a straight line. Find a) ∠ AFC, b) ∠ BEC.

*Solution:*

CDF is an equilateral triangle

DF = DC = FC

∠ DFC = ∠ FCD = ∠ FDC = 60^{0}

ABCD is a square

∠ ADC = 90^{0}

∠ ADF = 90^{0} – 60^{0} = 30^{0}

For a square

AD = DC

AD = DC = DF, ADF form an isosceles triangle

∠ AFD = ∠ DAF

∠ AFD = (180^{0} – 30^{0}) ÷ 2 = 75^{0}

∠ AFC = 60^{0 }+ 75^{0} = 135^{0}

BGC is a triangle

∠ GBC + ∠ GCB = 180^{0} – 105^{0} = 75^{0}

BEC is a triangle

∠ BEC = 180^{0} – ∠ GBC – ∠ GCB – 17^{0} – 18^{0}

∠ BEC = 180^{0} – (∠ GBC + ∠ GCB) – 35^{0}

∠ BEC = 180^{0} – 75^{0} – 35^{0} = 70^{0}