If x is an angle in standard position with point A(-3 , 4) on the terminal side, then sec(x) =
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calendarEducator since 2010
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We have that x is an angle in standard position with point A(-3 , 4) on the terminal side.
sec x = 1/ cos x
cos x = adjacent side / hypotenuse.
Here the adjacent side is -3 and the hypotenuse is sqrt ((-3^2) + 4^2) = sqrt ( 9 + 16) = sqrt 25 = 5.
The value of cos x = -3/5
Therefore the required value of sec x is -5/3
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calendarEducator since 2008
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We will draw a line between the point (-3, 4) and the origin point.
Then angle x located between the line and the x-axis.
Then, we hace formed a right angle triangle where the base = 3, the height = 4 .
Now we will calculate the length of the line which is the hypotenuse.
==> The hypotenuse = sqrt(3^2 + 4^2 ) = sqrt25 = 5
Now we will determine sec(x).
We know that sec(x) = 1/cos(x)
But cos(x) = adjacent / hypotenuse.
==> cos(x) = 3/5
But the angle is in the 4th quadrant.
then cos(x) = -3/5
==> sec(x) = -5/3
The terminal point A has the coordinates (-3, 4).
Let O be the origin whose coordinates are (0,0).
x is the angle the point OA makes with x axis positive direction.
Therefore OA = sqrt{(-3-0)^2+(4-0)^2)} = 5.
Let the feet of perpendicular from A to x axis be X.
Then OX = -3, XA = 4.
Then angle AOX = x.
Therefore secant x = hypotenuse/ (adj side to angle x) = OA/OX
= 5/(-3) = -5/3.
So sec(x) = -5/3.
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