Maharashtra State BoardHSC Science (Electronics) 11th
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Answer the following: Find the trigonometric functions of : −150° - Mathematics and Statistics

Diagram
Sum

Answer the following:

Find the trigonometric functions of :

−150°

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Solution

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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APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 2 Trigonometry - 1
Miscellaneous Exercise 2 | Q II. (1) | Page 33
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