Draw An Angle In Standard Position

SKETCH AN ANGLE IN STANDARD POSITION IF TERMINAL ARM PASSES THROUGH

Sketch an Angle in Standard Position if Terminal arm Passes Through :

Here we are going to see how to sketch an angle in standard position if the terminal arm passes through the given point.

Question 1 :

Sketch an angle in standard position so that the terminal arm passes through each point.

(a) (2, 6)

Solution :

First let us plot the point (2, 6) in the graph paper.

(b) (-4, 2)

(d) (-1, 0)

Determine the Trigonometric Ratios of the Given Angles - Examples

Question 1 :

Determine the exact values of the sine, cosine, and tangent ratios for each angle

Solution :

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

OA = Hypotenuse side

OB = Adjacent side and AB = Opposite side

sin 60 = √3/2

cos 60 = 1 /2

tan 60 = √3

Since the terminal arm lies in first quadrant, we have to take positive sign for sin 60.

(ii)

Solution :

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

<BOA = 180 + θ

180 + θ = 225

θ = 225 - 180 = 45

Since the terminal arm lies in the 3rd quadrant, we have to use positive sign for tan and cot only.

sin 225 = -1 / √ 2

cos 225 = -1 / √ 2

tan 225 = 1

(iii)

Solution :

By drawing a perpendicular line from A, we get a triangle OAB.

In triangle OAB,

<AOB = 180 - θ

<AOB = 180 - 150

<AOB = 30

sin 150 = 1/ 2

cos 150 = - √3/2

tan 150 = -1/ √3

(iv)

Solution :

sin 90 = 1

cos 90 = 0

tan 90 = undefined

Question 3 :

The coordinates of a point P on the terminal arm of each angle are shown. Write the exact trigonometric ratios sin θ, cos θ, and tan θ for each.

Solution :

Since the terminal arm lies in 3rd quadrant, we have to take positive signs for all trigonometric ratios.

Horizontal length = 3

Vertical length = 4

Hypotenuse =√3² + 4²

Hypotenuse = 5

sinθ = 4/5

cosθ = 3/5

tanθ = 4/3

(ii)

Solution :

Since the terminal arm lies in 3rd quadrant, we have to take positive sign only for tan and cot.

Horizontal length = 12

Vertical length = 5

Hypotenuse =√12² + 5²

Hypotenuse = 13

sinθ = -5/13

cosθ = -12/13

tanθ = 5/12

(iii)

Since the terminal arm lies in 4th quadrant, we have to take positive sign only for cos and sec.

Horizontal length = 8

Vertical length = 15

Hypotenuse =√8² + 15²

Hypotenuse = 17

sinθ = -15/17

cosθ = 8/17

tan θ = -15/8

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