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प्रश्न
Answer the following:
Find the trigonometric functions of :
−150°
उत्तर
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 3rd 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)`
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