मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Answer the following: Find the trigonometric functions of : −150° - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Answer the following:

Find the trigonometric functions of :

−150°

आकृती
बेरीज

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)`

shaalaa.com
Trigonometric Functions of Specific Angles
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [पृष्ठ ३३]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
पाठ 2 Trigonometry - 1
MISCELLANEOUS EXERCISE - 2 | Q II) 1) viii) | पृष्ठ ३३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×