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Mathematics 18 Online
OpenStudy (anonymous):

Checking another question. Pythagorean Theorem.

OpenStudy (anonymous):

I need a good grade, so I'm going to check every question.

OpenStudy (anonymous):

There are some pretty easy relations for 30-60-90 triangles. Do you remember them at all?

OpenStudy (anonymous):

No.

OpenStudy (anonymous):

The "short leg" (across from the 30 degree angle) is half that of the hypotenuse. The "long leg" (across from the 60 degree angle) is half the hypotenuse multiplied by the square root of 3. Alternatively, you can use just one of these, and then the fact that \[a^2 + b^2 = c^2\] for a right triangle.

OpenStudy (anonymous):

Thanks for the info.

OpenStudy (anonymous):

A perhaps easier way to remember the "long side" is that it is the "short side" multiplied by root 3. \[Long Side = Short Side * \sqrt{3}\]

OpenStudy (anonymous):

OK thanks. When I get useful information, I save it as a bookmark, so I'm going to save this question as a bookmark.

OpenStudy (mathmale):

I think the drawing of this triangle is intentionally "not to scale," to throw you off tract. It's not necessary, but you might want to redraw it, so that the angles really look like 30, 60 and 90 degrees. Vandreigan is referring to the "special triangle," otherwise known as the 30-60-90 degree triangle, whose sides have lengths that are relatively easy to remember. Before I move on, would you be willing to try drawing the given triangle to scale? I do not mean in the least to undercut what Vandreigen is telling you; in fact, I've learned from his presentation the fact that one of the legs of this triangle is Sqrt(3) times the other leg.

OpenStudy (anonymous):

I'll try to draw the triangle to somewhat scale.

OpenStudy (anonymous):

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