Respuesta :
the angle of 5π/6 is 150 degree. start from the positive side of the x axis, rotate counterclockwise for 150 degree, the terminal side is in the second quadrant, forming a 30 degree angle with the negative side of the x axis. This 30 degree angle is what we need to focus on.
Draw a vertical line from your terminal point straight down to the negative side of the x axis, and draw another line from the terminal point to the center of the unit circle. Now you have a right triangle.
In a unit circle, the radius, that is, the hypotenuse of of this right triangle, is always 1.
sin30=y:r =1/2, so y is 1/2
cos30=x:r=√3/2, so x is √3/2
because the terminal point is in the second quadrant, x is negative, y is positive, therefore, the standard position of the intersection is at (-√3/2, 1/2)
Draw a vertical line from your terminal point straight down to the negative side of the x axis, and draw another line from the terminal point to the center of the unit circle. Now you have a right triangle.
In a unit circle, the radius, that is, the hypotenuse of of this right triangle, is always 1.
sin30=y:r =1/2, so y is 1/2
cos30=x:r=√3/2, so x is √3/2
because the terminal point is in the second quadrant, x is negative, y is positive, therefore, the standard position of the intersection is at (-√3/2, 1/2)
This question is based on the concept of circle. Therefore, the terminal point is in the second quadrant, x is negative, y is positive, Thus, the standard position of the intersection is at [tex]\bold{(\dfrac{-\sqrt{3} }{2} ,\dfrac{1}{2} )}[/tex].
Given:
The angle 5π/6 in standard position intersect the unit circle.
We need to determined the the point where the terminal side of the angle 5π/6 in standard position intersect the unit circle.
According to question,
The angle of 5π/6 is 150 degree. Start from the positive side of the x axis, rotate counterclockwise for 150 degree.
The terminal side is in the second quadrant, forming a 30 degree angle with the negative side of the x axis. This 30 degree angle is what we need to focus on.
Now, draw a vertical line from your terminal point straight down to the negative side of the x axis.
Draw another line from the terminal point to the center of the unit circle. Now, you have a right triangle.
In a unit circle, the radius, the hypotenuse of of this right triangle, is always,
[tex]sin \; 30^0 = \dfrac{y}{r} =\dfrac{1}{2} , \; y \; is \dfrac{1}{2} \\\\cos\;30^0=\dfrac{x}{r} =\dfrac{\sqrt{3} }{2} , \; x \; is \; \dfrac{\sqrt{3} }{2}[/tex]
Therefore, the terminal point is in the second quadrant, x is negative, y is positive, Thus, the standard position of the intersection is at [tex]\bold{(\dfrac{-\sqrt{3} }{2} ,\dfrac{1}{2} )}[/tex].
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https://brainly.com/question/15116222
