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Mathematics 17 Online
OpenStudy (anonymous):

In a 30-60-90 triangle, the hypotenuse is 20 feet long. Find the length of the length of the long leg and the short leg

OpenStudy (anonymous):

short - 10 long - \[10\sqrt{3}\]

OpenStudy (anonymous):

short leg is half of the hypotenuse

OpenStudy (anonymous):

:))

OpenStudy (sriram):

20cos60=10 and 20sin60=10 *(30)^1/2

OpenStudy (anonymous):

@siriam: LOLwut?

OpenStudy (anonymous):

You can just use some formulas because the given is a special triangle.

OpenStudy (anonymous):

you need to trig for this. short side is half of long side because it is half of an equilateral triangle then the other side comes from pythagoras.

OpenStudy (anonymous):

i meant you need NO trig

OpenStudy (sriram):

dude 20/short side=cos60 & 20 / long side=sin60 whats wrong in it??

OpenStudy (anonymous):

there's nothing wrong with it, it's just the harder option.

OpenStudy (anonymous):

Nothing. You can just use simple equations for special triangles.

OpenStudy (anonymous):

nothing wrong except that the only reason you know the cosine of 60 degrees is because you know the ratios, not the other way around!

OpenStudy (anonymous):

short=20*sin30=10 long=20*sin60=10\[\sqrt{3}\]

OpenStudy (anonymous):

Why are you applying trigonometry on such easy questions?

OpenStudy (anonymous):

You can just use simple equations on special triangles.

OpenStudy (anonymous):

in a 30-60-90, a = short side, b = long side, c = hypotenuse c = 2a ; b = \[ a \sqrt{3}\]

OpenStudy (anonymous):

i think u did it too.i have no problem with trigonometry

OpenStudy (anonymous):

\[\text{long} \text{side}=20\text{Cos}\left[\text{ArcSin}\left[\frac{10}{20}\right]\right]=10 \sqrt{3} \]\[20^2=10^2+\left(10 \sqrt{3}\right)^2, 400=100+300, 400= 400 \]

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