Subcategories

• Day 5

  Trapezoid Lengths

  Module 2 Day 5 Challenge Part 1 Module 2 Day 5 Challenge Part 2 Module 2 Day 5 Challenge Part 3 Module 2 Day 5 Challenge Part 4 Module 2 Day 5 Challenge Part 5 Module 2 Day 5 Your Turn Part 1 Module 2 Day 5 Your Turn Part 2 Module 2 Day 5 Your Turn Part 3
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  @debbie

  How do we know that the two horizontal lines are parallel? They looked parallel but the question does not tell us it is.

• Day 6

  Angles in Circles

  Module 2 Day 6 Challenge Part 1 Module 2 Day 6 Challenge Part 2 Module 2 Day 6 Challenge Part 3 Module 2 Day 6 Challenge Part 4 Module 2 Day 6 Your Turn Part 1 Module 2 Day 6 Your Turn Part 2
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  @amusingmarmoset yes it is pretty easy, u just isosceles triangle ur way to the answer

• Day 7

  Chord Distances

  Module 2 Day 7 Challenge Part 1 Module 2 Day 7 Challenge Part 2 Module 2 Day 7 Challenge Part 3 Module 2 Day 7 Challenge Part 4 Module 2 Day 7 Challenge Part 5 Module 2 Day 7 Your Turn Part 1 Module 2 Day 7 Your Turn Part 2
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  @imadandelion No problem! 😄

• Day 8

  Cyclic Quadrilateral

  Module 2 Day 8 Challenge Part 1 Module 2 Day 8 Challenge Part 2 Module 2 Day 8 Challenge Part 3 Module 2 Day 8 Challenge Part 4 Module 2 Day 8 Your Turn Part 1 Module 2 Day 8 Your Turn Part 2
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  @victorioussheep Good questions! The "two angles that are missing" are these two, marked in yellow, which do turn out to be 60 degrees (I think Prof. Loh also points to them in the video, but I see how it could be a bit unclear).
  missingangles.png

  Solving for those yellow angles, rather than the red ones you marked, is a good way to use what you know about inscribed angles and cyclic quadrilaterals. Specifically, Prof. Loh uses the fact that an inscribed angle's measure is half the measure of the arc it cuts off.

  But you could definitely solve for the red angles first! Take the triangle with angles 20, 60, and (unknown red angle). The unknown red angle is equal to 100 degrees because the angles of a triangle add to 180. Then, the supplementary red angle is 80 degrees (because 100+80 = 180). Finally, the last red angle equals 180-40-80 = 60 degrees (using the triangle with angle 40, 80, (unknown red/blue angle)), like we found before.

  So basically, there are multiple valid ways to solve the problem, and each way emphasizes different methods 🙂 finding the yellow angles first is most useful for illustrating the usefulness of inscribed angles, but you could solve it either way.