Six-Trig Ratio│Please Help!

Question

Given the point (-6, 12) determine the six-trig ratio. Determine the two complementary angles. Label the angles G, H, I, with H being the right angle.

Answer 1

(-6, 12)=(x,y)=(a,o)

h=sqr(35+144)=6sqr5

sinG = o/h= 2/sqr5 cscG = h/o= .5sqr5

cosG = a/h= -6/sqr5 secG= h/a= -(sqr5)/6

tanG=o/a= -2 cotG= -.5

G=arcsin(2/sqr5)= about 63.43 degrees

complementary angle = 90- arcsin(2/sqr5)= about 26.57

Answer 2

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Answer 3

l (-6, 12)

l

l

I

l

l

H (-6,0)-------------- G (0,0)

This is the second quadrant. x < 0, y > 0

HG = x = -6

HI = y = 12

IH = Hyp = h = root (12^2 + (-6)^2 ) = root (180) = 6 sqrt (5)

sin G = O / h = y / h = 12 / 6 sqrt (5) = 2 / sqrt (5) = 2 sqrt (5) / 5

cos G = a / h = x / h = -6 / 6 sqrt (5) = -1 / sqrt (5) = - sqrt (5) / 5

tan G = o / a = y / x = 12 / -6 = -2

csc G = 1 / sin G = sqrt (5) / 2

sec G = sqrt (5) / -1 = - sqrt (5)

cot G = 1 / -2 = -1 / 2

m < G = arcsin (2 sqrt (5) / 5) = 63.43494882 ~~ 63.435 deg

m < I = 90 - 63.435 = 26.565 deg