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OpenStudy (anonymous):
Given the triangle below, what is sec∡B?
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OpenStudy (anonymous):
OpenStudy (anonymous):
A
OpenStudy (anonymous):
B
OpenStudy (anonymous):
C
OpenStudy (anonymous):
D
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OpenStudy (dumbcow):
C
OpenStudy (anonymous):
you serious?
OpenStudy (dumbcow):
umm yeah why wouldn't i be sec = 1/cos cos = sqrt(1-sin^2) = sqrt(1-(4/7)^2) = sqrt(1-16/49) = sqrt(33)/7 --> sec = 7/sqrt(33) = 7sqrt(33)/33
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