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katlin95
Group Title
please please help:) medals rewarded
find x
a. 8
b. 8sqrt3
c. 16 sqrt3
d. 1.8 sqrt3
 one year ago
 one year ago
katlin95 Group Title
please please help:) medals rewarded find x a. 8 b. 8sqrt3 c. 16 sqrt3 d. 1.8 sqrt3
 one year ago
 one year ago

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geoffb Group TitleBest ResponseYou've already chosen the best response.1
You have an angle, as well as hypotenuse and the opposite leg. Which function do you use when you have opposite and hypotenuse: sine, cosine, or tangent?
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
this is geometry
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Okay, then you need to know the rules of a 306090 triangle.
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Do you know how the three legs of a triangle relate?
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
not reaally but kinda
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
The ratio between short leg : long leg : hypotenuse is \(1 : \sqrt{3} : 2\).
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
Knowing that, you can use Pythagoras to solve for \(x\). Being across from the 60º angle, would it be the long leg or the short leg?
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
so wouldnt the answer be B to this question?
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
That's what I get, yes, since \(\large x = \frac{\sqrt{3}}{2}(h)\)
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
wouldnt it be B?
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
As I said, yes. \[\large x = \frac{\sqrt{3}}{2}(h) = \frac{16\sqrt{3}}{2} = 8\sqrt{3}\]
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
oh sorry their was a freezing problem! thanks so much:)
 one year ago

geoffb Group TitleBest ResponseYou've already chosen the best response.1
No problem. It's being slow on my side, too. Have a good day. :)
 one year ago

katlin95 Group TitleBest ResponseYou've already chosen the best response.0
yes you to!!!:)
 one year ago
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