SKETCH AN ANGLE IN STANDARD POSITION IF TERMINAL ARM PASSES THROUGH  A POINT

Example 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)

(c) (-5, -2) 

(d) (-1, 0)

Example 2 :

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)  

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)

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)

sin 90  =  1

cos 90 = 0

tan 90 = undefined

Example 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

sin θ  =  4/5

cos θ  =  3/5

tan θ  =  4/3

(ii)

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

sin θ  =  -5/13

cos θ  =  -12/13

tan θ  =  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

sin θ  =  -15/17

cos θ  =  8/17

tan θ  =  -15/8

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