Asked by Anonymous
P(-5, 5) is a point on the terminal side of θ in standard position. What is the exact value of sec θ?
Answers
Answered by
Reiny
your are in quad II
drop a perpendicular to the x-axis and draw your right-angled triangle
cosØ = x/r
secØ = r/x
remember : r^2 = x^2 + y^2
= (-5)^2 + 5^2
= 50
r = 5√2
secØ = 5√2/-5 = -√2
drop a perpendicular to the x-axis and draw your right-angled triangle
cosØ = x/r
secØ = r/x
remember : r^2 = x^2 + y^2
= (-5)^2 + 5^2
= 50
r = 5√2
secØ = 5√2/-5 = -√2
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