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OpenStudy (anonymous):
Find cos θ if csc θ = 2 and θ is in quadrant II.
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OpenStudy (anonymous):
OK so...
Sin θ = 1/Csc θ so it would be 1/2 which is O/H?
So the mesurements can be calculated using x^2 + y^2 = r^2?
OpenStudy (anonymous):
θ=60
OpenStudy (anonymous):
as......... if ....secθ=2
hence ...1/cosθ=2
cosθ=1/2
sin(90+θ)=1/2
OpenStudy (anonymous):
hence .......θ=60
OpenStudy (dumbcow):
csc = 1/sin
--> 1/sin = 2
--> sin = 1/2
cos^2 = 1-sin^2
cos^2 = 1-(1/2)^2
cos^2 = 3/4
cos = sqrt3/2
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OpenStudy (anonymous):
thought since sin would = 1/2 and csc = 2 then wouldn't it be 30?
OpenStudy (dumbcow):
no theta = 150 since in quad 2
OpenStudy (anonymous):
of it's cosec θ then dumbcow is right
OpenStudy (anonymous):
yea sqrt(3)/2
OpenStudy (anonymous):
would it be negative since it's in the II quadrant?
and only thing positive in II quadrant is Sin and CSC?
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OpenStudy (anonymous):
sry fakshon ...i consideed it as sec θ
OpenStudy (dumbcow):
correct, cos would be negative :)
OpenStudy (dumbcow):
@shivamsinha
it happens :)
OpenStudy (anonymous):
yep yep it's all good.
OpenStudy (anonymous):
thanks for the help.
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