Question

Find the trigonometric function of : 

150°

Answer 1

Trigonometric Functions of 150° : 

Let measure of ∠XOA in standard position be 150°.
Its terminal arm (ray OA) intersects the standard unit circle in P (x, y), which lies in the second quadrant.

Draw segment PM perpendicular to the X-axis.

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

Now, ΔOMP is a right-angled triangle in which m∠MOP  = 30° and OP = 1.

∴ MP = `1/2"OP" = 1/2 xx 1 = 1/2`

∴ | y | = `1/2`

By the distance formula,

x² + y² = 1

∴ `x^2 + (1/2)^2` = 1

∴ `x^2 + 1/4` = 1

∴ x² = `1 - 1/4`

∴ x² = `3/4`

∴ x = `+-(sqrt(3))/2 and y = ± 1/2`

But P lies in the second quadrant.

∴ x < 0 and y > 0

∴ x = `(-sqrt(3))/2, y = 1/2`

∴ P is `((-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)`