Question 480124: find cot b and sin b of an angle whose equation of its terminal side is 5x + 2y = 0 where x>0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find cot b and sin b of an angle whose equation of its terminal side is 5x + 2y = 0 where x>0
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5x+2y=0
rewrite to standard form of straight line: y=mx+b, m=slope,b=y-intercept
2y=-5x
y=-5x/2
This is an equation of a straight line with slope=-5/2 and y-intercept=0, which places the terminal side of reference angle b in quadrants II or IV.
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In quadrant II:
tan b=-5/2
cot b=-2/5
sin b=5/√29 (note that cos b is negative in this quadrant)
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In quadrant IV:
tan b=-5/2
cot b=-2/5
sin b=-5/√29 (note that cos b is positive in this quadrant)
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