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Mathematics 16 Online
OpenStudy (anonymous):

help

OpenStudy (anonymous):

OpenStudy (jdoe0001):

have you covered the so-called "special triangles" of 30-60-90 yet? what is the sum of all internal angles in a triangle anyway? notice you're given \(\cfrac{\pi}{3}\) and notice the "perpendicular segments" that means that angle is 90 degrees or \(\cfrac{\pi}{2}\)

OpenStudy (anonymous):

I know angle A is pi/6=60 degrees, but I don't know how to find length of a and b

OpenStudy (anonymous):

I mean pi/6 = 30degrees

OpenStudy (jdoe0001):

well... you're right A is \(\cfrac{\pi}{6}\) well.. so we have \(\begin{cases} \cfrac{\pi}{3}\to 60^o\\ \cfrac{\pi}{2}\to 90^o\\\ \cfrac{\pi}{6}\to 30^o \end{cases}\implies 30-60-90 \ rule\) notice the 30-60-90 rule -> http://www.gradeamathhelp.com/image-files/30-60-90-triangle.jpg notice the ratios keep in mind that we know that c = 9 or |dw:1413329195225:dw|

OpenStudy (jdoe0001):

so if you solve 9 = 2x for "x" then you can pretty much tell the other sides

OpenStudy (anonymous):

wow, you're right. My math teacher didn't talk about "special triangles" of 30-60-90" in class

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