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

Answer the following: Find the trigonometric functions of : 240° - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Answer the following:

Find the trigonometric functions of :

240°

आकृती
बेरीज

उत्तर

Trigonometric Functions of 240°:

Let the measure of ∠XOA in the standard position be 240°.

Its terminal arm (ray OA) intersects the standard unit circle in P (x, y), which lies in the third quadrant.

Draw segment PM perpendicular to the X-axis.

Then OM= l x l and MP = l y l.

In right-angled triangle OMP, m∠MOP = 60° and OP = 1

∴ m∠MOP = 30°

∴ OM = `1/2"OP" = 1/2 xx 1 = 1/2`

∴ | x | = `1/2`

By the distance formula,

x2+ y2= 1

∴ `(1/2)^2 + y^2` = 1

∴ `1/4 + y^2` = 1

∴ y2 = `1 - 1/4 = 3/4`

∴ y = `± sqrt(3)/2 and x = ±1/2`

But P lies in the third quadrant

∴ x < 0 and y < 0

∴ x = `-1/2 and y = -sqrt(3)/2`

∴ P is `(-1/2, -sqrt(3)/2)`

∴ sin 240° = y = `-sqrt(3)/2`

cos 240° = x = `-1/2`

tan 240° = x = `y/x = ((-(sqrt(3))/2))/((-(1)/2)) = sqrt(3)`

cosec 240° = `1/y = 1/((-(sqrt(3))/2)) = -2/sqrt(3)`

sec 240° = `1/x = 1/((-(1)/2))` = – 2

cot 240° = `x/y = ((-(1)/2))/((-(sqrt(3))/2)) = 1/sqrt(3)`

shaalaa.com
Trigonometric Functions with the Help of a Circle
  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) iv) | पृष्ठ ३३

संबंधित प्रश्‍न

Answer the following:

Find the trigonometric functions of :

90°


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×