Subcategories

• Module 2 Day 6 Challenge Part 1

  Isosceles Triangles, Radii

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  @amusingmarmoset yes it is pretty easy, u just isosceles triangle ur way to the answer

• Module 2 Day 6 Challenge Part 2

  Exterior Angle Theorem, Central Angles

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  central-angle-same-as-arc.png

  central-angle-same-as-arc-examples.png

• Module 2 Day 6 Challenge Part 3

  Formula for Inscribed Angle

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  potato art baby

• Module 2 Day 6 Challenge Part 4

  Proof of Formula for Inscribed Angle, Arc Measure

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  @mirthfulostrich Hi, thanks for your reply! The reason Prof. Loh made that \(\textcolor{green}{\text{green}}\) line is because it's a diameter. Why is that nice? A diameter goes through the center, so this way we get two radii. Why are the radii nice to have? It's because we can then construct isosceles triangles which have two radii as their sides. Why is it nice to have isosceles triangles? It's because isosceles triangles have two equal angles, and we're trying to prove something about the angles in this drawing, in particular, the fact that

   
  $$ \text{ An inscribed angle is half the measure of the central angle} $$

  (If you're interested in this proof, here is a visual proof of the Inscribed Angle Theorem, and here are some examples of inscribed angles.)

   
  Step by step, here is what Prof. Loh did for his proof:

  ) We want to show that the angle marked with a \( \textcolor{orange}{\text{yellow}}\) arrow is half of the measure of the central angle (the one formed by the radii meeting at the center).

  temp-20200909-1.png

  2.) Draw the \(\textcolor{green}{\text{green}}\) diameter, and label the angle formed by the \(\textcolor{green}{\text{green}}\) line and the left-hand radius \(\angle a.\) The triangle with two radii (marked by the yellow ticks) as sides is isosceles, so the \(\textcolor{purple}{\text{purple}}\) angles are the same size.

  temp-20200909-2.png

  3.) Using the Exterior Angle Theorem, we know the \(\textcolor{purple}{\text{purple}}\) angles add up to \(\angle a,\) so each of them is equal to \( \angle \frac{1}{2}a.\)

  temp-20200909-3.png

  4.) Remember, we actually want the angle that was marked \( \textcolor{orange}{\text{yellow}}\) above, so, in order to get that, we need to label some other angles. What other information can we use in the diagram? Well, there is another isosceles triangle in the picture! It's the one with two other radii as sides, outlined in \(\textcolor{red}{\text{red}}\) below:

  temp-20200909-4.png

  5.) One of the larger angles of this triangle is the angle that we want, but first, we need to find some other angle that relates to this triangle. That's why Prof. Loh made the angle \( \angle b,\) labeled in \(\textcolor{pink}{\text{pink}}.\) This (\textcolor{pink}{\text{pink}}\) angle is an exterior angle of the \(\textcolor{red}{\text{red}}\) triangle. Using the Exterior Angle Theorem again, each of the larger angles of this triangle is equal to half of the exterior angle, so each is \( \angle \frac{1}{2} b.\)

  temp-20200909-5.png

  6.) The end is in sight! The angle that we want is the \(\textcolor{pink}{\text{pink}}\) angle, \(\angle \frac{1}{2} b,\) minus the \(\angle \frac{1}{2} a \) angle, or

  $$ \text{ Angle we want } = \frac{1}{2} (b-a) $$

  7.) What were we trying to prove again? We were trying to show that an inscribed angle is half of the central angle. (Remember that these angles are inscribing the same arc of the circle.) In the case of our diagram, we were trying to show that the \( \textcolor{orange}{\text{yellow}}\) angle is half of the \(\textcolor{green}{\text{green}}\) angle.

  temp-20200909-7.png

  Well, what is the \(\textcolor{green}{\text{green}}\) angle? It is equal to \( \angle b - \angle a.\)
  And since the \( \textcolor{orange}{\text{yellow}}\) angle is \( \frac{1}{2} (b-a),\) it is indeed true that the \( \textcolor{orange}{\text{yellow}}\) angle is half of the \(\textcolor{green}{\text{green}}\) angle. Hooray, so does this prove the Inscribed Angle Theorem? Well, yes! Even if you draw the lines at different angles, these relationships still hold.

• Module 2 Day 6 Your Turn Part 1

  Point to Circle Line Segments, Derivation of Angle Formula

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  @The-Blade-Dancer You are right, using the Inscribed Angle Theorem would work a lot faster. 🙂 That theorem tells us that we can take the difference of the two arcs that are inscribed by angle \(P,\) and divide that by \(2\) to get the measure of \(\angle P.\)

   

  Screenshot from 2020-11-12 17-15-19.png

  $$ \angle P = \frac{80^{\circ} - 30^{\circ}}{2} = \frac{50^{\circ}}{2} = \boxed{25^{\circ}} $$

  You've finished the course already and are just reviewing, and so this theorem, which is proved in Day 10, isn't something that the kids know about when they're doing the Day 6 lesson. Maybe that's why Prof. Loh tried to use other methods to solve it! 🙂 That's my guess, but I also know that Prof. Loh likes to use different solution methods to check his work, also. 🙂

• Module 2 Day 6 Your Turn Part 2

  Secant Lines to a Circle, Angles of Intersecting Chords

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  @The-Rogue-Blade

  Ah! Good question! It's because you want to look for the "X" in the diagram. The "X" will tell you which two angles go together in the equation given in the video.

  M2W2D6-y-part-2-triple-chords-solution-why-not-80-a-and-b.png
   

  The arcs \((80 - a)\) and \(a\) are both inscribed by the "X." This is why we found that the angle \(40^{\circ}\) was the average of \(a\) and \(80-a.\)

   
  M2W2D6-y-part-2-triple-chords-solution-10-percent.png

   
   
  However, the arc \(b\) and the arc \((80-a)\) aren't inscribed by two chords which intersect in an "X."
   

  M2W2D6-y-part-2-triple-chords-solution-not-80-a-and-b.png

  Instead, the arc \(b\) pairs with the arc \(c,\) to form an "X." Thus, the following would be true:

  $$ \frac{b + c}{2} = 70^{\circ} $$

  🙂