if cos(𝜃) = 1/6 what is csc(90° − 𝜃)

Answer 1

cscT = 1/sinT

sinT = opposite side over hypotenuse

cosT = adjacent side over hypotenuse = 1/6

o^2 = h^2 -a^2 = 36-1 = 35

o= sqr35 = opposite side

sinT = sqr35/6,

but

sin(90-T) = cosT = 1/6

csc(90-T) = 1/sin(90-T) = 1/(1/6)

csc(90-T) = exactly 6

check the answer, find angle T

cosT= 1/6, T = about 80.4 degrees

sinT = sin80.4 = about .986

sin(90-80.4) = sin9.6 = about .167

csc(90-80.4) = about 1/.167 = about 5.996

csc(90-T) = 6

that's one Quadrant I solution for T and 90-T

T also = about 80.4 + 360n degrees where n= any integer

there are an infinite number of Quadrant I and IV solutions for T

but csc(90-T) = 6 for all solutions of T in Quadrant I

and -6 in Quadrant IV