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OpenStudy (anonymous):
Find x in these 30 - 60 - 90 triangles.
https://media.glynlyon.com/g_geo_2012/11/groupi39.gif
x = ?
It's in the form of sqrt.
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OpenStudy (anonymous):
is it trigono ?
OpenStudy (anonymous):
yeah, kinda. It's geometry, but they're using the sin,cos, and tan.
OpenStudy (anonymous):
@tanjung
OpenStudy (anonymous):
it would be a special triangle
OpenStudy (anonymous):
you would write
30 : 60 :90
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OpenStudy (anonymous):
then find what is across the 30, 60 , 90
OpenStudy (anonymous):
get me?
OpenStudy (anonymous):
30:60:90
x: 7:y
1:square root 3: 2
OpenStudy (jdoe0001):
based on the ratios on the 30-60-90 rule
\(\huge 7 = x\sqrt{3}\)
then just solve for "x" :)
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OpenStudy (anonymous):
No, I don't get it..
OpenStudy (anonymous):
This is how they want it:
OpenStudy (jdoe0001):
hmm, well
$$
7 = x\sqrt{3} \implies x = \frac{7}{\sqrt{3}}\\
\text{now in simplified form}\\
\implies \cfrac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \implies \cfrac{7\sqrt{3}}{(\sqrt{3})^2}\\
\cfrac{7\sqrt{3}}{3}
$$
OpenStudy (anonymous):
Thanks @jdoe0001
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