Let’s take a look at the problem that this student was struggling with.

Because of their relatively high difficulty compared to other questions, I felt it was appropriate to address an example of how I’d go about solving one in a blog post.

In my opinion, these are some of the hardest questions that the SAT tests you on. More specifically, they were struggling with Right Triangle Trigonometry questions. We can divide 78 by 5 to clear the fraction and get the decimal 15.Recently, I’ve been tutoring for the SAT, and I noticed one of my students was struggling with some of the SAT math questions. We need a Common Denominator of 5, so we multiply 13/1 (13 as a fraction) by 5: Since we're trying to find AC, we add 13 + 13/5.

So we'll take sides DE corresponding to BE, because we have the measure of both. 'Next, we'll set up proportions to solve for the length of EC.' That means take two corresponding sides of each triangle and set them equal to each other, and one of those sides will be the one we're trying to find, which is EC. 'Then, you'll need to recognize that the smaller triangle is similar' to the bigger triangle and '△AED is similar to △CEB' Okay, not too bad. Square Root of both sides (to clear the x^2) makes 13 = x (Because sq rt of 169 = 13)Īlso, you could remember that 5-12-13 is a common Pythagorean Triple.Īnyway, that makes the Hypotenuse of the big triangle AE = 13.

Oh man - that's a heck of a question! Okay let's see.Īlright so 'Since we're dealing with right triangles, you'll need to use the Pythagorean theorem to figure out the missing side length of the big triangle'.

Sat math khan academy