Answer 1

Angle of measure (– 150°) :

Let m∠XOA = – 150°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis

∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

OM = `sqrt(3)/2"OP"`

= `sqrt(3)/2(1)`

= `sqrt(3)/2`

PM = `1/2"OP"`

= `1/2(1)`

= `1/2`

Since point P lies in the 3^rd quadrant,

x < 0, y < 0

∴ x = – OM = `-sqrt(3)/2 and y = -"PM" = -1/2`

∴ P ≡ `(-sqrt(3)/2, (-1)/2)`

sin (– 150°) = y = `-1/2`

cos (– 150°) = x = `(-sqrt(3))/2`

tan (– 150°) = `y/x = (-1/2)/(-sqrt(3)/2) = 1/sqrt(3)`

cosec (– 150°) = `1/y = 1/((-1/2))` = – 2

sec (– 150°) = `1/x = 1/((-sqrt(3)/2)) = -2/sqrt(3)`

cot (– 150°) = `x/y = (-sqrt(3)/2)/(-1/2) = sqrt(3)`