What has been your most effective method for teaching to solve for two missing sides of a 45-45-90 and a 30-60-90? I have found that teaching it with formulas like a, a(root2) and a, a(root3), and 2a tends to confuse many kids due to introducing a third variable (in the formulas) when often there are already two variables in a diagram. I have tried teaching it through words or rules. Ex: “the medium side is equal to the short side times root(3).” Kids tend to just forget these rules or mix them up. Eventually, I have settled on simply teaching the 1:1:root(2) ratio and the 1:root(3):2 ratio for side lengths, and then the kids set up proportions using these and the corresponding values from the diagram. It is more mechanical than other intuitive approaches, but I think the kids who excel with the intuitive approaches will find their own mental or visual shortcuts anyway. The only thing I dislike about this is that it can take longer than more intuitive approaches. But, just making up numbers here, 80% of kids doing it successfully in a longer way is better than 60% doing it correctly in a shorter way. We will see what the data says come assessment time. What have been other math teachers’ experiences with this topic and the various ways to teach it?

While I have not taught that level, I certainly have utilized this tons (Bachelor's degree in math, write math competition tests annually). Truthfully, I utilize ratios as I do it...always remembering that it's a right triangle, so it better follow the Pythagorean theorem, and knowing that the angle opposite the 30 degree angle will be the smaller one of the two (for the 45/45/90, clearly the two legs are equal). This makes it so I can "recreate" the ratios with ease whenever I need.

45-45-90: a=b c=a*sqrt(2) 30-60-90: a = shorter leg (opposite 30) Think: Smallest angle is opposite the smallest side. b = longer leg (opposite 60) c = hypotenuse a=a b=a*sqrt(3) c=2a

I wish I saved my plans. When first approaching special triangles, students should be able to design/experiment/measure with either a computer aided program like Sketchpad, or easily programmable TI-nspire. 30-60-90 can be started with multiple sized equilateral triangles, students or programs can measure those sides. Let the students bisect, if able, or fold, or cut all the triangles. Measure, measure, measure. Construct a table to share results. Start with a square for the other one, same thing.....

I always just had them label s, s*sqrt(3), and 2*s on the triangle (s standing for short side), and then solve for s based on the side they were given. After solving the sides using just s, then I had them deal with whatever x or y is. I used similar with l, l, l*sqrt(2).

When you go over Pythagorean Theorem prior to this, do you have your students always treat the short leg as a and the long leg as b? I have always told mine it didn’t matter which leg was a or b (since it generally doesn’t), but maybe doing it this way would be a natural transition into keeping everything in terms of a, b, and c for your formulas. The only thing close to this I have done in prior years is: Leg = a Hyp = a*sqrt(2) Short Leg = a Long Leg = a*sqrt(3) Hyp = 2a But I looked for another way because many kids got confused since this introduced another variable in addition to the x and y that were usually already there.

I usually tell them ignore x and y entirely until the very end of the problem. If they want to cross out x and y to start, go for it! Then bring x and y back into the mix at the very last instance.

That’s a good idea. In any case, I’ve got a year to think on it. Already taught it one way this year which works decently, so no need to try to introduce a different method now and confuse everyone.

Technically speaking, it doesn’t matter which variable you use because you obtain the same result, which is why it is called a dummy variable. However, the students need something that they can easily remember, which is why I teach it that way. It works for the vast majority of my students.

Draw the generic 30-60-90 or 45-45-90 right triangle next to triangle you are trying to solve Use AA similarity and write a proportion to find the missing side length of the given triangle in comparison to it's generic triangle. x/2 = 2/1 Using cross products property 1*x = 2*2 x = 4

Reading all those numbers/words and math things........is really making my brain hurt. This is why I teach mid-elementary, where instead of triangle number algebra thing, we have find the angle, the basics of the distributive property, and identifying angles.

The variables are arbitrary! Teach your students how to develop the ratios/relationships themselves. It will take more than once it takes longer in the middle school grades and should be a short review in high school. ie., entry task. Given an equilateral triangle, develop the ratios for the side lengths of a 30-60-90 triangle.

Sorry if that it appears to be odd for me to have this much issue with something that is evidently straightforward, however English and remote dialect were my best subjects, and, to me, learning English/Language Arts takes almost no work. I experience difficulty identifying with my understudies who battle, and I don't know how to decipher the showing methodologies I've learned into something that will fit the ELA classroom.