Plsssssss helppppp

Question

Solve for x and y pls. Link to math problem

Answer 1

Hi Emma A.,

This, and the previous, and the following questions you submitted have a unifying theme: 1) Find the "special triangle"; 2) Use formulas (or in your head) to solve for the various unknown side lengths.

There are two type of special angles, that will sometimes make the entire triangle special. They are: 45 degrees, and 30 degrees. If there is a right triangle (one with a 90 degree angle) somewhere in the figure, OR if you can draw a 90 degree angle so as to enable a special triangle, you are most of the way to a soltion.

It would profit you greatly to memorize the side ratios of a 45/45/90 (degrees of the three angles) and a 30/60/90 triangle. These are respectively 1:1:sqrt(2) for the first triangle there, and 1:sqrt(3):2 for the second triangle. When you have a special triangle, first draw it as a sketch and label your sketch with the ratio numbers I just gave you. Then go into your actual figure, and scale up the sketch side lengths to match whatever is given you in the figure.

So, for example, perhaps you have a 45/45/90 triangle, and one of the two identical sides is given as sqrt(2). Insidious of your teacher!. But you draw your sketch, and say, "that is one of the length 1 sides in my sketch". "Therefore, I'm scaling everything up by sqrt(2)", and the hypotenuse of the given triangle is sqrt(2)*sqrt(2) = 2 in length.

I'm counting on you to apply this information on your other problems, which I'm marking as "identical topic answered".

-- Cheers, --Mr. d.