# 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  =  3Vertical length  =  4Hypotenuse  =  √32 + 42Hypotenuse  =  5sin θ  =  4/5cos θ  =  3/5tan 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  =  12Vertical length  =  5Hypotenuse  =  √122 + 52Hypotenuse  =  13sin θ  =  -5/13cos θ  =  -12/13tan 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  =  8Vertical length  =  15Hypotenuse  =  √82 + 152Hypotenuse  =  17sin θ  =  -15/17cos θ  =  8/17tan θ  =  -15/8

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