• M2 M5★

    Module 2 Week 3 Day 12 Your Turn Part 2

    I THOUGHT PO SHEN LOH SAID "ANY OLD 30-60-90" RIGHT TRIANGLE. SO WHY IS THE ANSWER TO TO DAY 13 YOUR TURN PART 2 A TRIANGLE THAT ISN'T A 30 60 90. SOMEHOW IT'S A 5 12 13. WOW MAGIC! HOPE YOU HEAR THE SARCASM!!!

  • ADMIN M0★ M1 M5

    @bravekiwi Yes, you are right that at the very end, Prof. Loh says, "This question shows us what happens when you take any old 30-60-90 right triangle and draw a line though the center." Even though the mini-question at the end of the Your Turn (Part 2) video seems like a very different triangle, it's actually illustrating the same concept as in the original Your Turn question! We could even use a random triangle to illustrate this concept, which has a nice name: it's called the Angle Bisector Theorem.

    When you have a line bisecting an angle in a triangle, that line will cut the opposite side in two parts, the ratios of which are in the same ratio as the other two sides of the triangle!

    M2W4D13-y-part-2-angle-bisector-solution1.png

    So the ratio of the green and purple sides, \(13:12,\) is the same as the ratio of the segments cut from the third side, \(13:12.\)

    In fact, if we had used another \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, then it might not have been obvious that this fact works for any triangle! It might have appeared to be a property of only \(30^{\circ}-60^{\circ}-90^{\circ}\) right triangles. 🙂

  • M0★ M1★ M2★ M3★ M4 M5

    Where did we get 13a and 12a?

  • ADMIN M0★ M1 M5

    @TSS-Graviser Hi there, the \(13a\) and \(12a\) come from the fact that we know the ratios of the \( \textcolor{green}{\text{green}}\) and \(\textcolor{blue}{\text{blue}}\) sides. How do we know the ratio of the \( \textcolor{green}{\text{green}}\) and \(\textcolor{blue}{\text{blue}}\) sides? From the information given in the question, that the yellow segment is 52% of the vertical side.

     

    temp-m2d13-y-2-screenshot.png

     

    Since \(52%\) equals the fraction \(\frac{52}{100},\) which equals \(\frac{4 \times 13}{4 \times 25},\) which reduces to \(\frac{13}{25},\) this means that the other piece of the vertical line under the yellow arrow is \(1 - \frac{13}{25} = \frac{12}{25} \) of the whole vertical side.

    That's how we got this picture here:

    M2W4D13-y-part-2-angle-bisector-solution1-50-percent.png

     

    Do you see that Angle Bisector Theorem here tells us that the ratio of the \( \textcolor{green}{\text{green}}\) and \(\textcolor{blue}{\text{blue}}\) sides is \( 13 : 12\) ?

    We can't say for sure that the \( \textcolor{green}{\text{green}}\) side is \(13\) long and the \(\textcolor{blue}{\text{blue}}\) side is \(12\) long, but we just know that the length of the \( \textcolor{green}{\text{green}}\) side divided by the length of the \(\textcolor{blue}{\text{blue}}\) side is equal to \(\frac{13}{12},\) so we say that the \( \textcolor{green}{\text{green}}\) side has length \(13\) units of some amount, which is \(13a,\) and the \(\textcolor{blue}{\text{blue}}\) side has length \(12\) units of some amount, which is \(12a.\)