Mathematics
9 Online
OpenStudy (anonymous):
sec 45 degrees=?
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OpenStudy (anonymous):
one over the cosine of 45 degrees
OpenStudy (jhannybean):
\[\sec(x) = \frac{1}{\cos(x)}\]
OpenStudy (solomonzelman):
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\(\large\color{slate}{ \sec(45^\circ)=s\sqrt{2}/s=? }\)
OpenStudy (anonymous):
do you know what
\[\cos(45^\circ)\] is ?
OpenStudy (solomonzelman):
guess that is a better start sate...
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OpenStudy (jhannybean):
Okay i am bad at maaking the little degree symbol -.-
OpenStudy (anonymous):
\circ
OpenStudy (anonymous):
\[x^\circ\]
OpenStudy (jhannybean):
\[\cos(45^\circ) =\frac{1}{\cos(45^\circ)}\]
OpenStudy (jhannybean):
there we go.
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OpenStudy (anonymous):
nice
OpenStudy (anonymous):
cos 45=1/1\[\sqrt{2}\] on a 45 45 90 triangle
OpenStudy (anonymous):
i menat 1/1 square root of 2
OpenStudy (anonymous):
i think that was supposed to be \(\frac{1}{\sqrt2}\) right?
OpenStudy (anonymous):
yeah
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OpenStudy (anonymous):
flip it
OpenStudy (anonymous):
square root of 2
OpenStudy (anonymous):
yup
OpenStudy (jhannybean):
\[\cos(45^\circ) =\frac{1}{\cos(45^\circ)} = \frac{1}{\dfrac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}=\sqrt{2}\]
OpenStudy (jhannybean):
Good job.
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OpenStudy (anonymous):
thank you so much! and this will apply to a 45 45 90 triangle right?
OpenStudy (anonymous):
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