हिंदी

Answer the following: Find the trigonometric functions of : −210° - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Answer the following:

Find the trigonometric functions of :

−210°

आकृति
योग

उत्तर

Angle of measure (– 210°) :

Let m∠XOA = −210°

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

PM = `1/2"OP"`

= `1/2(1)`

= `1/2`

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

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

= `sqrt(3)/2`

Since point P lies in the 2nd quadrant,

x < 0, y > 0

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

∴ P ≡ `(- sqrt(3)/2, 1/2)`

sin (−210°) = y = `1/2`

cos (−210°) = x = `(-sqrt(3))/2`

tan (−210°) = `y/x = ((1/2))/(((-sqrt(3))/2)) = - 1/sqrt(3)`

cosec (−210°) = `1/y = 1/((1/2))` = 2

sec (−210°) = `1/x = 1/(((-sqrt(3))/2)) = - 2/sqrt(3)`

cot (−210°) = `x/y = (((-sqrt(3))/2))/((1/2)) = -sqrt(3)`

shaalaa.com
Trigonometric Functions of Specific Angles
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Trigonometry - 1 - MISCELLANEOUS EXERCISE - 2 [पृष्ठ ३३]

APPEARS IN

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

Englishहिंदीमराठी


      Forgot password?
Use app×