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Find the trigonometric function of : 30° - Mathematics and Statistics

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प्रश्न

Find the trigonometric function of :

30°

आकृति
योग

उत्तर

Angle of measure 30°:


Let m∠XOA = 30°

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 1st quadrant,

x > 0, y > 0

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

∴ P = `(sqrt(3)/2, 1/2)`

sin 30° = y = `1/2`

cos 30° = x = `sqrt(3)/2`

tan 30° = `y/x = (1/2)/(sqrt(3)/2) = 1/sqrt(3)`

cosec 30° = `1/y = 1/((1/2))` = 2

sec 30° = `1/x=1/(((sqrt3)/2))=2/sqrt3`

cot 30° = `x/y = (sqrt(3)/2)/(1/2)=sqrt3`

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Trigonometric Functions of Specific Angles
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Trigonometry - 1 - EXERCISE 2.1 [पृष्ठ २१]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 2 Trigonometry - 1
EXERCISE 2.1 | Q 1) ii) | पृष्ठ २१

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